This is a good question because it points out thing people say ("Maxwell's equations are linear") which is true but can be easily interpreted incorrectly.
First, just to clarify, a set of differential equations for functions $F_i(\vec{x}, t)$ are linear if they have the form
$$
\sum_j \hat{D}_{ij}(\vec{x}, t) F_j(\vec{x}, t) = J_i(\vec{x}, t)
$$
where $\hat{D}_{ij}(x)$ is some set of differential operators that can depend on $\vec{x}$ and $t$ but not on $F$, and $J_i$ is a set of known functions. If $J_i=0$, this implies the superposition principle, that if $F_i(\vec{x}, t)$ and $G_i(\vec{x}, t)$ are solutions, then so is $a F_i + b G_i$ for constants $a$ and $b$. If $J_i$ is non-zero, then there's still a version of superposition; if $F_i$ is a solution for source $J_i$, and $G_i$ is a solution for source $K_i$, then $F_i+G_i$ is a solution with source $J_i+K_i$.
For fixed charge density $\rho(\vec{x}, t)$ and current density $\vec{J}(\vec{x}, t)$, Maxwell's equations are linear differential equations for the electric field $\vec{E}$ and magnetic field $\vec{B}$ (provided the charge density and current density obey the continuity equation, $\partial_t \rho + \nabla \cdot \vec{J}=0$, otherwise the equations aren't consistent).
However, if you want to include the dynamics of charges and currents, then the resulting equations aren't linear in general. The term "Maxwell's equations" usually does not refer to this situation, it refers to the situation where the charges and currents are known, or maybe when they obey linear constitutive equations. However, this won't be the case if you have a bunch of electrons, and you want to see how they interact. You have given a good physical argument for why electrons coupled to the electromagnetic field can't be a linear system. An electron at rest produces a static field, whereas two electrons will repel, and two electrons repelling each other is clearly not a superposition of two static electrons.
The theory that describes electrons coupled to the electromagnetic field is usually studied in quantum field theory and is called quantum electrodynamics. If you worked out its equations of motion, you would find they are not linear.
You don't need to go to QED to see where the problem is, though. Just looking at the force between that point charge of charge $q_2$ at position $r_1$ exerts on a point charge of charge $q_1$ and position $\vec{r}_1$ (please double check me on the sign)
$$
\vec{F}_1 = k q_1 q_2 \frac{\vec{r}_1 - \vec{r}_2}{|\vec{r}_1-\vec{r}_2|^3}
$$
you can see that the differential equation $\vec{F}_1=m\ddot{\vec{r}}_1$ for particle $1$ will be non-linear in $\vec{r}_1(t)$.
Even Newtonian gravity has this type of non-linearity. In other words, Newtonian gravity is a linear differential equation for the Newtonian potential $\phi$ if the masses are known, but if you want to see how the masses plus the gravitational potential evolve, then you have a non-linear set of equations. (As an aside, the fact that Newtonian gravity with dynamical point masses is non-linear is the reason the famous three body problem is so hard!)
General Relativity, however, has an even deeper kind of non-linearity. Even if there was no matter at all, the equations would still be non-linear. This is not true of Maxwell's equations -- if $\rho=J=0$, then the equations for $E$ and $B$ are linear -- nor of Newtonian gravity -- if the mass is zero, then the gravitational force is zero. Einstein's equations, however, are non-linear in the metric tensor even when the stress-energy tensor $T_{\mu\nu}=0$. This is why GR is able to have non-trivial vacuum solutions, called black holes! An observer can experience gravitation in GR, even if the stress-energy tensor vanishes everywhere in the spacetime! One (somewhat sloppy) way to say it is that the gravitational field can act as its own source, because of the non-linearity -- gravity gravitates!
