The basic assumptions on the space-time structure in classical mechanics are:
(1) Time intervals beetween events are absolute.
(2) Space intervals beetween contemporary events are absolute.
We may refere to this two properties as to the "galilean space-time structure". In the first chapter of Arnold's "Mathematical Methods of Classical Mechanics" you can find a mathematical formulation of space-time structure, and one of the problems is to prove that
All the affine transformations of space time which preserve time intervals and distances beetween contemporary events are compositions of rotations, translations and uniform motions.
The principle of relativity alone allows a much wider class of transformations (for example, it allows Lorentz's transformations).
So I think that your second and third questions can be rephrased in this way:
Can the galilean structure of space-time be obtained from known equations of motions (and the principle of relativity)?
I'm not able to be cathegorical about this point, but my intuition says that the answer is no, without at least some other strong assumptions.
Here's a simple derivation of galilean transformations.
Let $(t,\mathbf r)$, $(t',\mathbf r'$), denote the coordinates of an event in two frames of reference (not necessarily inertial frames).
In first place, the invariance of time intervals beetween events implies that
$$t'=t+t_0,$$
for some constant $t_0$.
Now, the invariance of space intervals beetween simultaneous events implies that, for a fixed $t$, $\mathbf r \mapsto \mathbf r'$ is an isometry of $\mathbb R ^3$. The most general form of such isometry is: $$\mathbf r'= \mathbf s+G\mathbf r,$$
where $GG^T=I$ and both $\mathbf s$ and $G$ may depend on time.
To establish that $\dot {\mathbf s}=0$ and $\dot G = 0$ if the two frames are inertial we notice that, by the principle of relativity, the equation: $$\ddot {\mathbf r}=0$$ for an isolated body must be covariant; a direct calculation shows that this is possible only if the above conditions are satisfied.
I'm no expert on the historical development of the subject, however I will offer a derivation.
Consider two frames of reference $S$ and $S'$, and suppose that $S'$ moves with speed $\textbf v$ with respect to $S$. Coordinates in $S$ and $S'$ are related by a Galileian transformation:
$$\begin{cases} t' = t \\ \textbf x' = \textbf x-\textbf vt\end{cases}$$
To find how the fields transform, we note that a Lorentz transformation reduces to a Galileian transformation in the limit $c \to \infty$. In fact, under a Lorentz transformation the fields transform like:
$$ \begin{cases}
\textbf E' = \gamma (\textbf E + \textbf v \times \textbf B) - (\gamma-1) (\textbf E \cdot \hat{\textbf{v}}) \hat{\textbf{v}}\\
\textbf B' = \gamma \left(\textbf B - \frac{1}{c^2}\textbf v \times \textbf E \right) - (\gamma-1) (\textbf B \cdot \hat{\textbf{v}}) \hat{\textbf{v}}\\
\end{cases}$$
Taking the limit $c\to \infty$ so that $\gamma\to 1$, we obtain the Galileian transformations of the fields:
$$ \begin{cases}
\textbf E' = \textbf E + \textbf v \times \textbf B\\
\textbf B' = \textbf B\\
\end{cases}$$
We can then invert the transformation by sending $\textbf v \to -\textbf v$:
$$ \begin{cases}
\textbf E = \textbf E' - \textbf v \times \textbf B'\\
\textbf B = \textbf B'\\
\end{cases}$$
By the same reasoning, can obtain the Galileian transformation of the sources:
$$ \begin{cases}
\textbf J = \textbf J' + \rho' \textbf v\\
\rho = \rho'\\
\end{cases}$$
We know that the fields and sources satisfy Maxwell's equations in $S$:
$$ \begin{cases}
\nabla \cdot \textbf E = \rho/\epsilon_0\\
\nabla \cdot \textbf B = 0\\
\nabla \times \textbf E = -\frac{\partial \textbf B}{\partial t}\\
\nabla \times \textbf B = \mu_0 \left(\textbf J +\epsilon_0 \frac{\partial \textbf E}{\partial t} \right)\\
\end{cases}$$
Replacing the fields and sources in $S$ with those in $S'$ we obtain:
$$ \begin{cases}
\nabla \cdot \textbf (\textbf E' - \textbf v \times \textbf B') = \rho'/\epsilon_0\\
\nabla \cdot \textbf B' = 0\\
\nabla \times \textbf (\textbf E' - \textbf v \times \textbf B') = -\frac{\partial \textbf B'}{\partial t}\\
\nabla \times \textbf B' = \mu_0 \left(\textbf J' + \rho' \textbf v +\epsilon_0 \frac{\partial (\textbf E' - \textbf v \times \textbf B')}{\partial t} \right)\\
\end{cases}$$
As a last step, we need to replace derivatives in $S$ with derivatives in $S'$. We have:
$$\begin{cases} \nabla = \nabla' \\ \frac{\partial }{\partial t} = \frac{\partial }{\partial t'} - \textbf v \cdot \nabla\end{cases}$$
Substituting and removing the primes and using vector calculus, we obtain:
$$ \begin{cases}
\nabla \cdot \textbf E + \textbf v \cdot (\nabla \times \textbf B) = \rho/\epsilon_0\\
\nabla \cdot \textbf B = 0\\
\nabla \times \textbf E = -\frac{\partial \textbf B}{\partial t}\\
\nabla \times \textbf B = \mu_0 \left(\textbf J + \rho \textbf v +\epsilon_0 \frac{\partial}{\partial t}( \textbf E - \textbf v \times \textbf B) - \epsilon_0 \textbf v \cdot \nabla (\textbf E - \textbf v \times \textbf B) \right)\\
\end{cases}$$
In a vacuum, we can take the curl of the fourth equation to obtain:
$$c^2\nabla^2 \textbf B = \frac{\partial^2 \textbf B}{\partial t^2} + (\textbf v \cdot \nabla)^2 \textbf B - 2 \textbf v \cdot \nabla \left(\frac{\partial \textbf B}{\partial t}\right)$$
Substituting a wave solution of the form $\textbf B \sim \exp{i(\textbf k \cdot \textbf x -\omega t)}$
We obtain an equation for $\omega$, which we can solve to obtain:
$$\omega = -\textbf v \cdot \textbf k \pm c |\textbf k|$$
Therefore the speed of propagation is the group velocity:
$$\frac{\partial \omega}{\partial \textbf k} = -\textbf v \pm c \hat{\textbf{ k}}$$
which gives you the expected $c\pm v$ with an appropriate choice of $\textbf v$ and $\textbf k$.
Best Answer
Maxwell's equations have wave solutions that propagate with speed $c = \frac{1}{\sqrt{\mu_0\epsilon_0}}$.
Since velocity is relative (speed c with respect to what?), it was initially thought that the what is an luminiferous aether in which electromagnetic waves propagated and which singled out a family of coordinate systems at rest with respect to the aether.
If so, then light should obey the Galilean velocity addition law. That is, a lab with a non-zero speed relative to the luminiferous aether should find a directionally dependent speed of light.
However, the Michelson–Morley experiment (original and follow-ups) failed to detect such a directional dependence. Some implications are
(1) there is no aether and electromagnetic waves propagate at an invariant speed. This conflicts with Galilean relativity for which two observers in relative uniform motion will measure different speeds for the same electromagnetic wave. This path leads to special relativity theory.
(2) there is an aether but it is undetectable. This path leads to Lorentz aether theory.