First things, first: Lorentz's aether theory is not equivalent to Special Relativity! It leads to the wrong equations - the non-relativistic Maxwell-Thomson equations, rather than their relativistic version - the Maxwell-Minkowski equations. I'll address this in greater depth and - in the process - also answer your question.
Maxwell's equations can be divided into two parts. First, are the equations that are invariant under arbitrary coordinate transformations:
$$
= ∇×,\quad = -∇φ - \frac{∂}{∂t},\quad ∇· = 0,\quad ∇× + \frac{∂}{∂t} = ,\\
∇· = ρ,\quad ∇× - \frac{∂}{∂t} = ,\quad ∇· + \frac{∂ρ}{∂t} = 0,
$$
where
$$∇ = \left(\frac{∂}{∂x},\frac{∂}{∂y},\frac{∂}{∂z}\right).$$
Their general coordinate-invariance is made manifest, and shown, by writing them in the language of differential forms:
$$
dA = F,\quad dF = 0,\\
dG = J,\quad dJ = 0,
$$
with the differential forms given by:
$$
A = ·d - φdt,\quad F = ·d + ·d∧dt,\\
G = ·d - ·d∧dt,\quad J = ρdV - ·d∧dt,
$$
where
$$ = (x,y,z),\quad d = (dx, dy, dz), \quad d = (dy∧dz, dz∧dx, dx∧dy),\quad dV = dx∧dy∧dz.$$
I can't overemphasize that they literally are invariant under all coordinate transforms. The coordinates $(x,y,z,t)$ don't have to be Cartesian, in the equations above, but can be any four independent functions of the space-time coordinates. The respective components for the fields are then picked off from the differential forms, themselves. The resulting equations will have exactly the same form.
Second are the equations that break this symmetry - the constitutive relations. For a medium that has at least one frame in which the constitutive relations are isotropic, they have the following form:
$$ + α × = ε( + β ×),\quad - α × = μ( - β ×).$$
This is where we get to the crux of all the issues raised.
First, these equations are the form that are suitable for a geometry that has the following as its invariants:
$$
βdt^2 - α\left(dx^2 + dy^2 + dz^2\right),\\
β\left(\left(\frac{∂}{∂x}\right)^2 + \left(\frac{∂}{∂y}\right)^2 + \left(\frac{∂}{∂z}\right)^2\right) - α\left(\frac{∂}{∂t}\right)^2,
$$
and
$$dt\frac{∂}{∂t} + dx\frac{∂}{∂x} + dy\frac{∂}{∂y} + dz\frac{∂}{∂z},\tag{1}\label{1}.$$
The symmetries that leave these invariants fixed also leave the constitutive laws fixed. In infinitesimal form, they are:
$$Δ = × - βt + ,\quad Δt = -α· + τ,$$
where $(, , , τ)$ are, respectively, infinitesimal rotation, boost, spatial translation and time translation. For the coordinate differentials, the corresponding transforms are, determined by the condition $dΔ(\_) = Δd(\_)$ and are given by:
$$Δd = ×d - βdt,\quad Δdt = -α·d,$$
while for the differential operators, they are determined by the requirement that ($\ref{1}$) be invariant:
$$Δ(∇) = ×∇ + α\frac{∂}{∂t},\quad Δ\left(\frac{∂}{∂t}\right) = β·∇.$$
The general transforms, written in infinitesimal form, for the fields are determined by the requirement that the respective differential forms be invariant. They are
$$
Δ = × - αφ,\quad Δφ = -β·,\quad Δ = × - α×,\quad Δ = × + β×,\\
Δ = × + α×,\quad Δ = × - β×,\quad Δ = × - βρ,\quad Δρ = -α·,
$$
and (lest we forget):
$$Δ = × - + αβ·.$$
The vacuum corresponds to the case where $ε = ε_0$, $μ = μ_0$ and $βεμ = α$, in which case (provided that $αβ||^2 < 1$), the constitutive equations become independent of $$ and can just be written as $ = ε_0 $ and $ = μ_0$. This only leads to sensible results where $αβ > 0$. In the more general case, the Maxwell-Minkowski relations are for Relativistic moving media, and $$ is present.
For the non-relativistic case, $$ is mandatory. This extra velocity vector is what pre-1905 literature is referring to when drawing a distinction between the "stationary" and "moving" Maxwell equations, and is present in all the non-relativistic treatments, specifically including those by Lorentz, Hertz and Heaviside, not just Maxwell's. It is what the "moving" in the title of Einstein's 1905 "On the Electrodynamics Of Moving Bodies" is referring to.
Now ... to answer your question: the signature is determined here by the sign of the product $αβ$. If $αβ < 0$, then we have a definite - or Euclidean - signature. If $αβ > 0$, then we have an indefinite - or Minkowski - signature. If $αβ = 0$, then we have a singular signature. This can be further subdivided into the cases where $β ≠ 0$ - the Galilean signature (where $c = ∞$ and time is absolute), where $α ≠ 0$ - the Carrollean signature (where $c = 0$ and space is "absolute", i.e. there is an absolute rest frame) and where $(α,β) = (0,0)$, the "static" signature - which corresponds to the case where both space and time are absolute (i.e. where all speeds are absolute, as is simultaneity).
Finally ... to the matter of Lorentz and the fake news "Lorentz aether theory is equivalent to Special Relativity" meme: no, it's not. That's a myth. The equations for the constitutive law that Lorentz derived are equivalent to those corresponding to the case $α = 0$ and $β ≠ 0$, which is also equivalent to the forms written by Maxwell, after making a correction posed by Heaviside and Thomson, and written by Hertz. These are all non-relativistic - as is any theory that leads to these equations: including specifically, Lorentz's aether theory.
The equations with $αβ > 0$ are equivalent to those written by Einstein and Laub, and independently by Minkowski in 1907-1908 - the Maxwell-Minkowski equations, where light speed is given by $c = \sqrt{β/α}$.
Lorentz' equations are - up to a change in notation - identical to the non-relativistic case. This is the Rosetta Stone that lays out the correspondences between Lorentz' notation and the notation used above. I also put up one for Hertz, in one of my replies, which (if I hunt it down and find it) I will add as a comment. For Heaviside, I haven't written up a Rosetta Stone, yet.
