No, for one very simple reason: the equations he derived from his theory for electromagnetism are non-relativistic: i.e. the equations actually posed by Maxwell in his treatise with the correction by Heaviside and Thomson, rather than their relativistic counterparts: Maxwell-Minkowski equations.
I laid out a Rosetta Stone between Lorentz' Attempt of a Theory of Electrical and Optical Phenomena in Moving Bodies in my reply The Rosetta Stone For Lorentz showing that, in fact, his equations are just the Maxwell-(Heaviside-)Thomson equations, not the Maxwell-(Einstein-Laub-)Minkowski equations. Written in Abschnitt II of his article, Lorentz' equations are:
$$\begin{align}
Ⅰ_b\ & \text{Div}\ = ρ \\
Ⅱ_b\ & \text{Div}\ ℌ = 0 \\
Ⅲ_b\ & \text{Rot}\ ℌ' = 4πρ + 4π\dot{} \\
Ⅳ_b\ & \text{Rot}\ = -\dot{ℌ} \\
Ⅴ_b\ & = 4πV^2 + [·ℌ] \\
Ⅵ_b\ & ℌ' = ℌ - 4π[·] \\
Ⅶ_b\ & = + [·ℌ]
\end{align}$$
Reiterating what I noted in my reply about his notation, his "Div" is our $∇·(\_)$, his "Rot" is our $∇×(\_)$, where
$$∇ = \left(\frac{∂}{∂x}, \frac{∂}{∂y}, \frac{∂}{∂z}\right),$$
his $\dot{(\_)}$ is our (and his) $∂/∂t$, and his $[\_·\_]$ is our $(\_)×(\_)$, so that he could have just as well have written $[∇·\_]$ for "Rot". Strangely, now that I think about it, I don't recall seeing any instance of the vector dot product anywhere in his article! Otherwise, he could have written "Div" as $∇·(\_)$, alongside $[∇·\_]$ for "Rot". I will use the dot product, while retaining his notation elsewhere. So $(\_)·(\_)$ means our dot product, but $[(\_)·(\_)]$ means his cross product.
Finally, his $V$ is our $1/\sqrt{με}$; that is: the wave speed for the underlying medium. In vacuum, $V = 1/\sqrt{μ_0ε_0} = c$.
The Heaviside/Thomson correction is the $-4π[·]$ term in $Ⅵ_b$. Without it, Lorentz' equations are equivalent to the ones Maxwell actually listed in his treatise, which Maxwell went through pains to show have symmetry under the Galilean transform (or would - if you add in the Heaviside/Thomson correction - that was an oopsie on Maxwell's part).
By the way, the counterpart $(\_)_a$ equations are those for the "stationary frame" - i.e. the frame where $ = 0$. That, right there, is already a strong clue telling you that Lorentz is still couched in the world of non-relativistic physics, notwithstanding the mental gymnastics of his narrative.
The correct equations - when written in his notation - that accord with the theory of Relativity are the same as above except for $Ⅴ_b$ and $Ⅵ_b$, which have to be changed to:
$$\begin{align}
{Ⅴ_b}'\ & = 4πV^2 + [·ℌ] - \frac{V^2}{c^2}[·ℌ'], \\
{Ⅵ_b}'\ & ℌ' = ℌ - 4π[·] + \frac{[·]}{c^2},
\end{align}$$
which will make this equivalent to the equations laid out independently by Einstein & Laub, and by Minkowski around the same time in the 1907-1908 period - the Maxwell-Minkowski equations.
The other equations don't need to be changed ... because they are invariant under arbitrary coordinate transformations, which can be displayed in manifest form by writing them in the language of differential forms. Only $Ⅴ_b$ and $Ⅵ_b$ break this symmetry and make a distinction between the relativistic versus non-relativistic forms of Maxwell's Theory.
For the relativistic versions ${Ⅴ_b}'$ and ${Ⅵ_b}'$, the relations in a vacuum reduce to:
$$V = c,\quad = 4πc^2, \quad ℌ' = ℌ,$$
the vector $$ becomes "superfluous" (to use Einstein's term) and none of Lorentz' mental gymnastics are required. The distinction between the relativistic and non-relativistic versions can be put into sharp relief by using the parameter $α = 1/c^2$ for the relativistic case, and $α = 0$ for the non-relativistic case, writing equations $Ⅴ_b/{Ⅴ_b}'$ and $Ⅵ_b/{Ⅵ_b}'$ in combined form as:
$$\begin{align}
Ⅴ_{bα}\ & = 4πV^2 + [·ℌ] - αV^2[·ℌ'], \\
Ⅵ_{bα}\ & ℌ' = ℌ - 4π[·] + α[·].
\end{align}$$
These equations go with a space-time geometry that has the following as its invariants:
$$
dt^2 - α\left(dx^2 + dy^2 + dz^2\right),\\
dt \frac{∂}{∂t} + dx \frac{∂}{∂x} + dy \frac{∂}{∂y} + dz \frac{∂}{∂z},\\
\left(\frac{∂}{∂x}\right)^2 + \left(\frac{∂}{∂y}\right)^2 + \left(\frac{∂}{∂z}\right)^2 - α \left(\frac{∂}{∂t}\right)^2
$$
as its invariants.
Under a Galilean transform
$$ → - t,\quad t → t$$
(where $ = (x, y, z)$) we have the following transform:
$$
\dot{(\_)} → \dot{(\_)} + ·∇, \quad \text{Div} → \text{Div}, \quad \text{Rot} → \text{Rot}, \\
→ , \quad ℌ' → ℌ' - 4π[·],\\
ℌ → ℌ, \quad → + [·ℌ], \\
\quad ρ → ρ, \quad → , \\
\quad → - , \quad → + , \quad V → V$$
and - as you can clearly see - his equations are invariant under them; though $Ⅲ_b$ and $Ⅳ_b$ take a little vector algebra to see it and require application of $Ⅰ_b$ and $Ⅱ_b$, respectively.
To show that it's actually the modified equations that are Relativistic, rather than the equations Lorentz wrote down, it's easiest to set $α = 1/c^2$ and write the Lorentz transforms in infinitesimal form as:
$$Δ = -t,\quad Δt = -α·.$$
Then, under these transforms:
$$
Δ\dot{(\_)} = ·∇(\_),\quad Δ\text{Div} = α·\dot{(\_)},\quad Δ\text{Rot} → α\left[·\dot{(\_)}\right],\\
Δ = \frac{α\left[·ℌ'\right]}{4π},\quad Δℌ' = -4π[·],\\
Δℌ = α[·],\quad Δ = [·ℌ],\\
Δρ = -α·ρ, \quad Δ = α[·[·]],\\
Δ = - + α·,\quad Δ = - α·,\quad ΔV = 0,
$$
the modified equations will be invariant - but not Lorentz' equations.
In other words: Lorentz' equations are not invariant under the Lorentz transformations! It's the modified equations which are. His theory is non-relativistic.
Therefore, Lorentz' aether theory is not equivalent to Special Relativity and is - in fact - not even Relativistic at all, but non-Relativistic.
If he had been able to use his aether theory to derive the correct equations ${Ⅴ_b}'$, and ${Ⅵ_b}'$, instead of $Ⅴ_b$, and $Ⅵ_b$, then we could have said that Lorentz' theory was equivalent to Special Relativity. But he simply derived the wrong equations - the non-relativistic ones, and all of his mental gymnastics were for nought.
