I think I found the answer, it was a bit tricky so I am posting it - Using vector calculus identity we can start by expanding $\frac{\partial B}{\partial t}$:
$\frac{\partial B}{\partial t}$=$\nabla$x ($u'$x $B$) +$\nabla$x (($\Omega$ x $r$)x$B$$\nabla \times (u' \times B) +\nabla \times ((\Omega \times r)\times B$)
Here we need to use $\nabla\times\left(\mathbf{A}\times\mathbf{B}\right)=\mathbf{A}\left(\nabla\cdot\mathbf{B}\right)-\mathbf{B}\left(\nabla\cdot\mathbf{A}\right)+\left(\mathbf{B}\cdot\nabla\right)\mathbf{A}-\left(\mathbf{A}\cdot\nabla\right)\mathbf{B}$
So that $\nabla$x (($\Omega$ x $r$)x$B$) = $(\Omega \times r)(\nabla \cdot B)-B(\nabla \cdot (\Omega \times r))+(B \cdot \nabla)(\Omega \times r)-((\Omega \times r) \cdot \nabla)(B)$
Using the fact that $\nabla \cdot B=0$ and $\nabla \cdot (\Omega \times r)=0$ The two first terms are equal to zero
Another very useful vector trick is: $(B \cdot \nabla)(\Omega \times r)=\Omega \times (B \cdot \nabla)r = \Omega \times B$
We are left with: $\frac{\partial B}{\partial t}$=$\nabla$x ($u' \times B) +\Omega \times B -((\Omega \times r) \cdot \nabla)(B)$
Here $\frac{\partial B}{\partial t}$ is taken in the static frame, we need to convert that to the rotating frame: $\frac{\partial B}{\partial t}|_{static}= \frac{DB}{Dt}|_{static} - u_\Omega \cdot \nabla(B)$
to take in account advection as the velocity of the field with the frame is $u_\Omega=(\Omega \times r)$
$\frac{DB}{Dt}|_{static}=\frac{DB}{Dt}|_{rotating}+\Omega \times B$
But $\frac{DB}{Dt}|_{rotating}$ is identical to $\frac{\partial B}{\partial t}|_{rotating}$ because the frame is moving with $u_{\Omega}$ already.
Combining everything yields to: $\nabla \times (u' \times B) +\Omega \times B -((\Omega \times r) \cdot \nabla)(B)=\frac{\partial B}{\partial t}|_{rotating} + \Omega \times B - ((\Omega \times r) \cdot \nabla)(B)$
After simplifying - $\frac{\partial B}{\partial t}|_{rotating}=\nabla \times (u' \times B)$ ....
Now there is probably a physical argument that can spare me all the troubles but I really needed to see it mathematically. I think what was really tricky was to use the total derivative to be able to use the formula relating the derivatives in the different frames (/!\ that relation does not hold for partial derivatives)
Check that link for more info: http://physics.princeton.edu/~mcdonald/examples/rotatingEM.pdf
