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
You aren't "replacing a mathematical proof". What the statements you are referring to mean is that in tensor notation, the proof is immediate, so that nothing needs to be written down. This is because if you have a tensor equation as above, in order to prove Lorentz invariance, do a Lorentz transformation and go to another set of coordinates $x^{\mu'}$. Then using the usual transformation laws we get that ${\partial_{\mu}} = \Lambda^{\mu'}_{\mu}\partial_{\mu'}$ and $F_{\mu\nu} = \Lambda^{\mu'}_{\mu}\Lambda^{\nu'}_{\nu}F_{\mu'\nu'} $, we can write the Maxwell equation in terms of the new coordinates to become
$\Lambda^{\mu'}_{\mu}\Lambda^{\nu'}_{\nu}\Lambda^{\sigma'}_{\sigma}(F_{\mu'\nu',\sigma'}+F_{\nu'\sigma',\mu'}+F_{\sigma'\mu',\nu'})=0.$
However, this can only hold if the thing inside the brackets is zero itself. Namely Maxwell's equation in the primed coordinate system also holds.
More succintly, what a "tensor equation" means is that there was nothing special about the coordinate system in which the equations were derived. You could have equally well chosen another system and derived the same equations. Thus invariance under coordinate change is immediate.