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.
Galilean relativity is usually discussed in the context of Newtonian mechanics. The dynamics is governed by Newton's laws. Galilean relativity concerns kinematics and it says that the dynamical laws are covariant with respect to Galilean transformations. In other words, their form is invariant. You got that right.
Maybe it would be useful to look at it from a more abstract mathematical point of view.
In the Galilean worldview, spacetime is a 4-dimensional affine space $A^4$. Affine basically means that all the points are the same and you have to pick some point if you want to work in $\mathbb{R} \times \mathbb{R^3}$.
This is just saying that you have to pick the origin for your coordinate system in 3-dimensional space and some moment $t=0$ as the origin in time.
Next, you define your metrics because you want to be able to measure stuff. Spatial distance between two points in $\mathbb{R \times R^3}$ is defined as $$d(\mathbf{x}, \mathbf{y}) = \sqrt{\sum_{n=1}^{3} \left( y_n - x_n \right)^2}$$
Distance in time, i.e. the time interval is defined as $$\tau(\mathbf{x}, \mathbf{y}) = \left| y_0 - x_0 \right|$$ where the 0th component stand for time.
In this picture, Newton's 1st law simply means that inertial reference frames form an equivalence class of coordinate systems in $A^4$ with origins moving with constant velocity with respect to each other. More precisely, in an inertial reference frame, the distance to the origin of every other inertial reference frame is a linear function of time.
Newton's 2nd law tells us that any motion that isn't governed by a function linear in time, must have its motion determined by a force, a quantity (in general, a function of space and time) which must be proportional to the rate at which the body changes its velocity. The constant of proportionality is what we call mass.
Newton's 3rd law is of less relevance here, but let's just state it for the sake of completeness. It says that when one body is acting on another with some force, the second body is acting on the first in such a way that when added vectorially, the forces sum to zero.
Note how all these laws are in essence coordinate-free, you can understand them without having to imagine a coordinate system, because they talk about concepts which are invariant when your space has the Galilean structure. When writing down the equations, you have to introduce coordinates at some point if you want to get out some numbers out of the calculation, but the coordinates are simply a tool, something that's very common in mathematics, because we know how to work with numbers and how to do simple algebra and calculus.
To conclude, whenever you have Galileo-invariant quantities such as acceleration or force, those terms should stay exactly the same in all coordinate systems. These quantities depend only on the Galilean structure of spacetime.
On the other hand, Galileo-covariant quantities such as velocity and position will change, but in accordance with Galilean transformations. These quantities depend on your choice of coordinate systems and are not defined on $A^4$.
Best Answer
The Lorentz prediction and the Galilean prediction must agree in the limit that $v \to 0$ (or in the limit that $c \to \infty$). This is because $v=0$ corresponds to no transformation at all, so they had better both agree there. So if you take the transformation and evaluate it for smaller and smaller $v$, you'll find that $k=1$ still has to be true.
That's just the Galilean Transformation of the EM field. To see how it relates to the relativistic case, the Lorentz Transformation of the EM field is:
$$\mathbf{E}' = \gamma \left( \mathbf{E} + \mathbf{v} \times \mathbf{B} \right ) - \left ({\gamma-1} \right ) ( \mathbf{E} \cdot \mathbf{\hat{v}} ) \mathbf{\hat{v}}$$
$$\mathbf{B}' = \gamma \left( \mathbf{B} - \frac {\mathbf{v} \times \mathbf{E}}{c^2} \right ) - \left ({\gamma-1} \right ) ( \mathbf{B} \cdot \mathbf{\hat{v}} ) \mathbf{\hat{v}}$$
When you take the limit that $c \to \infty$, we know that $\gamma \to 1$, so it just becomes:
$$\mathbf{E}' = \mathbf{E} + \mathbf{v} \times \mathbf{B}$$
$$\mathbf{B}' = \mathbf{B}$$