$\boldsymbol{\S}$ A. The function $\,\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right)\,$ as a total derivative
We'll prove at an elementary level that for the electromagnetic field the Lorentz invariant scalar function $\,\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right)\,$ is the 4-divergence of a 4-dimensional vector function. So adding this scalar to the Lagrangian density of the field doesn't change the equations of motion, that is the Maxwell equations.
From the expressions of $\,\mathbf E,\mathbf B\,$ in terms of the scalar and vector potentials $\phi,\mathbf A\,$
\begin{align}
\mathbf E & \boldsymbol{=}\boldsymbol{-}\boldsymbol{\nabla}\phi\boldsymbol{-}\dfrac{\partial \mathbf A}{\partial t}
\tag{01a}\label{01a}\\
\mathbf B & \boldsymbol{=} \boldsymbol{\nabla}\boldsymbol{\times}\mathbf A
\tag{01b}\label{01b}
\end{align}
we have
\begin{equation}
\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right)\boldsymbol{=}\boldsymbol{-}\underbrace{\left(\boldsymbol{\nabla}\phi\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\left(\dfrac{\partial \mathbf A}{\partial t}\right)}}_{\boldsymbol{\boxed{\,1\,}}}\boldsymbol{-}\underbrace{\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)}_{\boldsymbol{\boxed{\,2\,}}}
\tag{02}\label{02}
\end{equation}
Our target would be to find, if there exist, a real scalar function $\,\eta\,$ and a real 3-vector function $\,\boldsymbol{\xi}\boldsymbol{=}\left(\xi^1,\xi^2,\xi^3\right)\,$ , that is a 4-dimensional vector function $\,\boldsymbol{\Xi}\boldsymbol{=}\left(\xi^1,\xi^2,\xi^3,\eta\right)\,$ such that it yields the equality
\begin{equation}
\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right)\boldsymbol{=}\dfrac{\partial \eta}{c\partial t}\boldsymbol{+}\boldsymbol{\nabla}\boldsymbol{\cdot}\boldsymbol{\xi}\boldsymbol{=}\partial_{\mu}\Xi^{\mu}
\tag{03}\label{03}
\end{equation}
In the following we'll make use of the identity
\begin{equation}
\boxed{\:\:\boldsymbol{\nabla}\boldsymbol{\cdot}\left(\mathbf a\boldsymbol{\times}\mathbf b\right)\boldsymbol{=}\mathbf b\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf a\right)\boldsymbol{-}\mathbf a\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf b\right)\vphantom{\dfrac{a}{b}}\:\:}
\tag{04}\label{04}
\end{equation}
Using the identity \eqref{04} with $\,\mathbf a\boldsymbol{\equiv}\boldsymbol{\nabla}\phi\,$ and $\,\mathbf b\boldsymbol{\equiv}\mathbf A\,$ we have
\begin{equation}
\boldsymbol{\boxed{\,1\,\:}}\boldsymbol{=}\left(\boldsymbol{\nabla}\phi\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\left(\dfrac{\partial \mathbf A}{\partial t}\right)}\boldsymbol{=}\mathbf A\boldsymbol{\cdot}\underbrace{\left[\boldsymbol{\nabla}\boldsymbol{\times} \left(\boldsymbol{\nabla}\phi\right)\vphantom{\dfrac{a}{b}}\right]}_{\boldsymbol{0}}\boldsymbol{-}\boldsymbol{\nabla}\boldsymbol{\cdot}\left[\left(\boldsymbol{\nabla}\phi\right)\boldsymbol{\times}\mathbf A \vphantom{\dfrac{a}{b}}\right]
\tag{05}\label{05}
\end{equation}
that is
\begin{equation}
\boldsymbol{\boxed{\,1\,\:}}\boldsymbol{=}\left(\boldsymbol{\nabla}\phi\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\left(\dfrac{\partial \mathbf A}{\partial t}\right)}\boldsymbol{=}\boldsymbol{\nabla}\boldsymbol{\cdot}\left[\boldsymbol{-}\left(\boldsymbol{\nabla}\phi\right)\boldsymbol{\times}\mathbf A \vphantom{\dfrac{a}{b}}\right]
\tag{06}\label{06}
\end{equation}
Now
\begin{equation}
\dfrac{\partial }{\partial t}\left[\mathbf A\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\dfrac{a}{b}}\right]\boldsymbol{=}\underbrace{\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)}_{\boldsymbol{\boxed{\,2\,}}}\boldsymbol{+}\mathbf A\boldsymbol{\cdot}\left[\boldsymbol{\nabla}\boldsymbol{\times}\left(\dfrac{\partial \mathbf A}{\partial t}\right)\right]
\tag{07}\label{07}
\end{equation}
From identity \eqref{04} with $\,\mathbf a\boldsymbol{\equiv}\left(\dfrac{\partial \mathbf A}{\partial t}\right)\,$ and $\,\mathbf b\boldsymbol{\equiv}\mathbf A\,$
\begin{equation}
\boldsymbol{\nabla}\boldsymbol{\cdot}\left[\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\times}\mathbf A\right]\boldsymbol{=}\mathbf A\boldsymbol{\cdot}\left[\boldsymbol{\nabla}\boldsymbol{\times}\left(\dfrac{\partial \mathbf A}{\partial t}\right)\right]\boldsymbol{-}\underbrace{\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)}_{\boldsymbol{\boxed{\,2\,}}}
\tag{08}\label{08}
\end{equation}
Subtracting equations \eqref{07},\eqref{08} side-by-side yields
\begin{equation}
\boldsymbol{\boxed{\,2\,\:}}\boldsymbol{=}\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\boldsymbol{=}\dfrac{\partial }{\partial t}\left[\frac12\mathbf A\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\dfrac{a}{b}}\right]\boldsymbol{-}\boldsymbol{\nabla}\boldsymbol{\cdot}\left[\frac12\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\times}\mathbf A\right]
\tag{09}\label{09}
\end{equation}
that is
\begin{equation}
\boldsymbol{\boxed{\,2\,\:}}\boldsymbol{=}\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\boldsymbol{=}\dfrac{\partial }{c\partial t}\left[\frac12 c \mathbf A\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\dfrac{a}{b}}\right]\boldsymbol{-}\boldsymbol{\nabla}\boldsymbol{\cdot}\left[\frac12\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\times}\mathbf A\right]
\tag{10}\label{10}
\end{equation}
From equations \eqref{02},\eqref{06} and \eqref{10} we have
\begin{equation}
\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right) \boldsymbol{=}\boldsymbol{-}\underbrace{\left(\boldsymbol{\nabla}\phi\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\left(\dfrac{\partial \mathbf A}{\partial t}\right)}}_{\boldsymbol{\boxed{\,1\,}}}\boldsymbol{-}\underbrace{\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)}_{\boldsymbol{\boxed{\,2\,}}}\qquad \boldsymbol{\Longrightarrow}
\nonumber
\end{equation}
\begin{equation}
\boxed{\:\:\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right)\boldsymbol{=} \dfrac{\partial }{c\partial t}\left[\boldsymbol{-}\frac12 c \mathbf A\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\dfrac{a}{b}}\right]\boldsymbol{+}\boldsymbol{\nabla}\boldsymbol{\cdot}\left[\frac12\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\times}\mathbf A\boldsymbol{+}\left(\boldsymbol{\nabla}\phi\right)\boldsymbol{\times}\mathbf A\right]\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:}
\tag{11}\label{11}
\end{equation}
So the Lorentz invariant scalar function $\,\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right)\,$ is the 4-divergence of the following 4-dimensional vector function
\begin{equation}
\boxed{\:\:\boldsymbol{\Xi} \boldsymbol{=}\left(\boldsymbol{\xi},\eta \right)\boldsymbol{=} \Biggl(\left[\frac12\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\times}\mathbf A\boldsymbol{+}\left(\boldsymbol{\nabla}\phi\right)\boldsymbol{\times}\mathbf A\right],\left[\boldsymbol{-}\frac12 c \mathbf A\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\dfrac{a}{b}}\right]\Biggr)\:\:}
\tag{12}\label{12}
\end{equation}
$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$
$\boldsymbol{\S}$ B. The function $\,\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right)\,$ as Lagrange density producing identically zero equations of motion
Motivated by John Dumancic's answer I give the proof of the above conclusion
So, consider that the Lagrangian density $\,\mathcal{L}$ is only this function
\begin{equation}
\mathcal{L}\boldsymbol{=}\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right)\boldsymbol{=}\left(\boldsymbol{-}\boldsymbol{\nabla}\phi\boldsymbol{-}\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)
\tag{C-01}\label{C-01}
\end{equation}
We must consider this density as function of the four ''field coordinates'', the components of the electromagnetic 4-vector
\begin{equation}
\mathcal A\boldsymbol{=}\left(A_0,A_1,A_2,A_3\right)\boldsymbol{=}\left(\phi,\mathbf A\right)
\tag{C-02}\label{C-02}
\end{equation}
and their time and space 1rst order derivatives so that
\begin{equation}
\mathcal{L}\left(A_{\jmath}, \dfrac{\partial A_{\jmath}}{\partial t}, \dfrac{\partial A_{\jmath}}{\partial x_k}\right) \qquad \left(\jmath=0,1,2,3\right) \qquad \left(k=1,2,3\right)
\tag{C-03}\label{C-03}
\end{equation}
We express the Lagrangian density of equation \eqref{C-01} in terms of these coordinates
\begin{align}
\mathcal{L}\boldsymbol{=}&\boldsymbol{-}\left(\dfrac{\partial \phi}{\partial x_1}\boldsymbol{+}\dfrac{\partial A_1}{\partial t}\right)\left(\dfrac{\partial A_3}{\partial x_2}\boldsymbol{-}\dfrac{\partial A_2}{\partial x_3}\right)\boldsymbol{-}
\left(\dfrac{\partial \phi}{\partial x_2}\boldsymbol{+}\dfrac{\partial A_2}{\partial t}\right)\left(\dfrac{\partial A_1}{\partial x_3}\boldsymbol{-}\dfrac{\partial A_3}{\partial x_1}\right)
\nonumber\\
&\boldsymbol{-}
\left(\dfrac{\partial \phi}{\partial x_3}\boldsymbol{+}\dfrac{\partial A_3}{\partial t}\right)\left(\dfrac{\partial A_2}{\partial x_1}\boldsymbol{-}\dfrac{\partial A_1}{\partial x_2}\right)
\tag{C-04}\label{C-04}
\end{align}
The Euler-Lagrange equations of motion are
\begin{equation}
\frac{\partial }{\partial t}\left[\frac{\partial \mathcal{L}}{\partial \left(\dfrac{\partial A_{\jmath}}{\partial t}\right)}\right]\boldsymbol{+}\sum_{k\boldsymbol{=}1}^{k\boldsymbol{=}3}\frac{\partial }{\partial x_{k}}\left[\frac{\partial \mathcal{L}}{\partial \left(\dfrac{\partial A_{\jmath}}{\partial x_{k}}\right)}\right]\boldsymbol{-} \frac{\partial \mathcal{L}}{\partial A_{\jmath}}\boldsymbol{=}0 \qquad \left(\jmath=0,1,2,3\right)
\tag{C-05}\label{C-05}
\end{equation}
For $\jmath\boldsymbol{=}0\:( A_0\boldsymbol{=}\phi)$ we have
\begin{equation}
\frac{\partial }{\partial t}\underbrace{\left(\frac{\partial \mathcal{L}}{\partial \overset{\:\,_{\boldsymbol \cdot}}{\phi}}\right)}_{0}\boldsymbol{+}\underbrace{\boldsymbol{\nabla}\boldsymbol{\cdot}\overbrace{\left(\frac{\partial \mathcal{L}}{\partial \boldsymbol{\nabla}\phi\vphantom{\overset{\:\,_{\boldsymbol \cdot}}{\phi}}}\right)}^{\boldsymbol{-}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)}}_{0}\boldsymbol{-} \underbrace{\frac{\partial \mathcal{L}}{\partial \phi\vphantom{\overset{\:\,_{\boldsymbol \cdot}}{\phi}}}}_{0}\boldsymbol{=}0
\tag{C-06}\label{C-06}
\end{equation}
that is the lhs is an identically zero term. This happens for the rest three equations, for example for $\jmath\boldsymbol{=}1$ we have
\begin{equation}
\underbrace{\frac{\partial }{\partial t}\overbrace{\left(\frac{\partial \mathcal{L}}{\partial \overset{\:\,_{\boldsymbol \cdot}}{A}_1}\right)}^{\dfrac{\partial A_2}{\partial x_3}\boldsymbol{-}\dfrac{\partial A_3}{\partial x_2}}}_{\dfrac{\partial^2 A_2}{\partial t\partial x_3}\boldsymbol{-}\dfrac{\partial^2 A_3}{\partial t\partial x_2}}\boldsymbol{+}\underbrace{\boldsymbol{\nabla}\boldsymbol{\cdot}\overbrace{\left(\frac{\partial \mathcal{L}}{\partial \boldsymbol{\nabla}A_1\vphantom{\overset{\:\,_{\boldsymbol \cdot}}{\phi}}}\right)}^{
\begin{bmatrix}
0\\
\dfrac{\partial \phi}{\partial x_3}\boldsymbol{+}\dfrac{\partial A_3}{\partial t}\\
\boldsymbol{-}\dfrac{\partial \phi}{\partial x_2}\boldsymbol{-}\dfrac{\partial A_2}{\partial t}
\end{bmatrix}
}}_{\dfrac{\partial^2 A_3}{\partial x_2\partial t}\boldsymbol{-}\dfrac{\partial^2 A_2}{\partial x_3\partial t}}\boldsymbol{-} \underbrace{\frac{\partial \mathcal{L}}{\partial A_1\vphantom{\overset{\:\,_{\boldsymbol \cdot}}{\phi}}}}_{0}\boldsymbol{=}0
\tag{C-07}\label{C-07}
\end{equation}
that is a lhs identically zero too. Similarly for $\jmath\boldsymbol{=}2,3$.
Conclusion : The function $\,\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right)\,$ as
Lagrangian density alone produces identically zero equations of motion. So, adding it to any Lagrangian density of the electromagnetic field has no effect to the equations of motion.
