I'll point out the more detailed differences below, but a nice rule of thumb to follow for these is that since the Galilean transformation gets it's name from a man who lived several centuries ago, the physics formulation for them is more basic than the Lorentz transformation, which is a more modern interpretation of physics. That way you can remember that the Galilean transformation is more of a crude approximation of the motion of particles, while Lorentz transformation are more exact.
Galilean Transformation
This is what most people's intuitive understanding of a particle in motion would be. Say we are inside a train that is moving. Now, if we start running across the train in the same direction, we would expect that to a viewer outside of the train, the speed of you running inside the train is the train's speed plus the speed at which you are running. However, this is not necessarily true. Back then, they expected this was a simple rule that followed from the laws of physics, but in more recent centuries it was proven not to be the case.
Below are the Galilean transformation equations. This is if we are moving in the x direction. We find that our position in another frame (not moving frame) is always given by our original position plus the velocity times time (the velocity of our moving frame, ie, the train). Notice that time in the first frame equals time in the second frame! This is not necessarily true! This is precisely why nobody really went against this transformation for 3 centuries or so. I mean who would have guessed that a guy inside a train measure time more slowly than a guy outside the train?
$x' = x - vt$
$y' = y$
$z' = y$
$t' = t$
Lorentz transformation
This gets a bit more complicated. Using the same example, it is proven that the speed that an outside observer watches you move is actually not your running speed plus the train's speed. This has to do with being in a moving reference frame when you, inside the moving train, measure your speed. To you, inside the train, you are the only thing doing the moving. You pretend the train is stationary, and the rest of the world is moving by you. Now an outside observer thinks that he/she is staionary, and that you are inside a moving reference frame. This brings up the principle of relativity.
Let's increase our speeds a lot to demonstrate the Lorentz transformation. Say the train is moving at .75c (.75 the speed of light) and then inside the train, you move at .5c. This would mean (using Galilean transformations) that an outside observer sees you moving at 1.25c! This is impossible, since Einstein tells us we can never move faster than the speed of light. Lorentz transformations take care of this paradox. To keep it sipmle, there is a scale factor that limits us from ever reaching a speed greater than the speed of light.
The key assumption that makes this possible is that time is not the same in all reference frames. This allows for length contraction, which allows for all of physics to still work while maintaining that all particles cannot exceed the speed of light. Below are the equations
$\beta = \frac{v}{c}$
$\gamma = \frac{1}{\sqrt{1 - \beta^2}}$
$t' = \gamma \left(t - \frac{vx}{c^2}\right)$
$x' = \gamma(x - vt)$
$y' = y$
$z' = z$
Let's look to your own statements.
First, time derivative after transformations isn't equal to an "old" derivative: for $\mathbf r' = \mathbf r - \mathbf u t = \mathbf r - \mathbf u t' \Rightarrow \mathbf r = \mathbf r' + \mathbf u t'$
$$
\partial_{t'} = (\partial_{t'}\mathbf r )\partial_{\mathbf r} + (\partial_{t'}t) \partial_{\mathbf t} = (\mathbf u \cdot \nabla ) + \partial_{t}, \quad (\mathbf u \cdot \nabla ) = u^{i}\partial_{x_{i}} .
$$
So, with $\nabla ' = \nabla$, "Bianchi" equations transforms to
$$
(\nabla \cdot \mathbf B') = 0 , \quad [\nabla \times \mathbf E '] + \frac{1}{c}\partial_{t}\mathbf B' + \frac{1}{c}(\mathbf u \cdot \nabla)\mathbf B ' = 0. \qquad (.1)
$$
Second, the form of $\mathbf {E}'(\mathbf r', t'), \mathbf B ' (\mathbf r ' , t')$ isn't equal to $\mathbf E (\mathbf r , t), \mathbf B (\mathbf r , t)$. Let's use the Lorentz force expression,
$$
\mathbf F = q\mathbf E + \frac{q}{c}[\mathbf v \times \mathbf B ].
$$
It doesn't depend on acceleration, so the statement that $\mathbf F ' = \mathbf F$ under galilean transformation is true. It means that
$$
\mathbf E + \frac{1}{c}[\mathbf v \times \mathbf B] = \mathbf E ' + \frac{1}{c}[\mathbf v ' \times \mathbf B'].
$$
By using galilean transformation for speed, $\mathbf v' = \mathbf v - \mathbf u$, this equation can be rewritten as
$$
\mathbf E + \frac{1}{c}[\mathbf v \times \mathbf B] = \mathbf E ' + \frac{1}{c}[\mathbf v \times \mathbf B '] - \frac{1}{c}[\mathbf u \times \mathbf B'], \qquad (.2)
$$
so the statement that $\mathbf E = \mathbf E ' , \quad \mathbf B = \mathbf B '$ isn't correct. So you need to find expressions $\mathbf E ' $ and $\mathbf B'$ via $\mathbf E $, $\mathbf B$.
By rewriting $(.2)$,
$$
\mathbf E + \frac{1}{c}[\mathbf v \times (\mathbf B - \mathbf B' )] = \mathbf E ' - \frac{1}{c}[\mathbf u \times \mathbf B '] ,
$$
in a reason of arbitrary $\mathbf u $ you can get the solution:
$$
\mathbf B' = \mathbf B , \quad \mathbf E' = \mathbf E + \frac{1}{c}[\mathbf u \times \mathbf B ].
$$
By substitution these equations to $(.1)$ you will get
$$
(\nabla \cdot \mathbf B)= 0 , \quad [\nabla \times \mathbf E] + \frac{1}{c}[\nabla \times [\mathbf u \times \mathbf B]] + \frac{1}{c}\partial_{t}\mathbf B + \frac{1}{c}(\mathbf u \cdot \nabla) \mathbf B = [\nabla \times \mathbf E] + \frac{1}{c}\partial_{t}\mathbf B = 0,
$$
because for $\mathbf u = const$
$$
[\nabla \times [\mathbf u \times \mathbf B]] = \mathbf u (\nabla \cdot \mathbf B) - (\mathbf u \cdot \nabla )\mathbf B = - (\mathbf u \cdot \nabla )\mathbf B .
$$
So the first pair of Maxwell's equations is clearly invariant under galilean transformations.
Let's look to the other pair of Maxwell's equations:
$$
[\nabla \times \mathbf B] - \frac{1}{c}\partial_{t}\mathbf E = 0 , \quad (\nabla \cdot \mathbf E ) = 0 . \qquad (.3)
$$
By using an expressions which were derived above, you can rewrite $(.3)$ as
$$
[\nabla \times \mathbf B] - \frac{1}{c}\partial_{t}\mathbf E ' - \frac{1}{c}(\mathbf u \cdot \nabla)\mathbf E' =
$$
$$
=[\nabla \times \mathbf B] - \frac{1}{c}\partial_{t}\mathbf E - \frac{1}{c}(\mathbf u \cdot \nabla)\mathbf E - \frac{1}{c^{2}}\partial_{t}[\mathbf u \times \mathbf B] - \frac{1}{c^{2}}(\mathbf u \cdot \nabla)[\mathbf u \times \mathbf B]= 0,
$$
$$
(\nabla \cdot \mathbf E ) + \frac{1}{c}(\nabla \cdot [\mathbf u \times \mathbf B]) = (\nabla \cdot \mathbf E ) -\frac{1}{c}(\mathbf u \cdot [\nabla \times \mathbf B]) = 0 .
$$
The requirement of galilean invariance of second equation leads to te state that $\frac{1}{c}(\mathbf u \cdot [\nabla \times \mathbf B])$, which isn't true in the general case. Analogically reasoning can be used for the first equation.
So the second pair of Maxwell's equations isn't invariant under Galilean transformations.
Best Answer
Your reasoning is spot on and the D'Alembert equation is indeed not Galilean-invariant: you are not missing anything aside from some historical knowledge; this is not my specialty either, but I think I can answer.
This non-Galilean-invariance was simply taken to be evidence for the existence of a luminiferous aether. The D'Alembert wave equation also perfectly well describes sound, and there is of course no problem with its non-Galilean-invariance here: this is exactly what you expect when there is a "privileged" frame defined by a wave's medium. When Maxwell's equations were discovered, the physics community simple assumed that they were correct only for the frame at rest relative to the luminiferous aether. In the middle of the 19th century, most researchers had abandoned Galileo's relativity postulate, at least for light. This was not an unreasonable stance until invalidated by experiments such as the Michelson-Morely experiment: Galileo knew nothing of electromagnetism or any details of physics at relative speeds comparable to the speed of light.