Instead of doing the $u$ substitution before evaluating the integral in $v$, it's easier to do it after computing that integral. Here is a detailed computation.
(I've assumed that you want to evaluate this integral and not simply to show that it is equal to $\zeta(2)$. This exercise appears in several references: Tom Apostol's A Proof that Euler Missed: Evaluating $\zeta(2)$ the Easy Way, Martin Aigner and Günter Ziegler's Proofs from The BOOK and as an exercise in a number theory text by LeVeque .)
By the substitution $x=\frac{\sqrt{2}}{2}\left( u-v\right) ,y=\frac{\sqrt{2}}{2}\left( u+v\right) $, whose Jacobian $J=\frac{\partial (x,y)}{\partial (u,v)}=1$, the region of integration becomes the blue square in the $u,v$-plane with vertices
$(0,0),\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right),(\sqrt{2},0),\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right), $ as shown in the following figure.

Observing that
$ \frac{1}{1-xy}=\frac{2}{2-u^{2}+v^{2}} $
is symmetric in $v$, we get
$\begin{eqnarray*} I &=&\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xy}\,dx\,dy & x=\frac{\sqrt{2}}{2 }\left( u-v\right) ,\quad y=\frac{\sqrt{2}}{2}\left( u+v\right) \\ &=&2\int_{u=0}^{\sqrt{2}/2}\int_{v=0}^{u}\frac{2}{2-u^{2}+v^{2}}\times 1\,du\,dv \\ &&+2\int_{u=\sqrt{2}/2}^{\sqrt{2}}\int_{v=0}^{\sqrt{2}-u}\frac{2}{ 2-u^{2}+v^{2}}\times 1\,du\,dv \\ &=&4\int_{0}^{\sqrt{2}/2}\left( \int_{0}^{u}\frac{dv}{2-u^{2}+v^{2}}\right) \,du\, \\ &&+4\int_{\sqrt{2}/2}^{\sqrt{2}}\left( \int_{0}^{\sqrt{2}-u}\frac{dv}{ 2-u^{2}+v^{2}}\right) \,du \\ &=&4\int_{0}^{\sqrt{2}/2}\frac{1}{\sqrt{2-u^{2}}}\arctan \frac{u}{\sqrt{ 2-u^{2}}}\,du, & u=\sqrt{2}\sin t \\ &&+4\int_{\sqrt{2}/2}^{\sqrt{2}}\frac{1}{\sqrt{2-u^{2}}}\arctan \frac{\sqrt{2 }-u}{\sqrt{2-u^{2}}}\,du\, & u=\sqrt{2}\cos \theta \\ &=&4\int_{0}^{\pi /6}\arctan \left( \tan t\right)\, dt\\&&+4\int_{0}^{\pi /3}\arctan \left( \frac{1-\cos \theta}{\sin \theta}\right) d\theta, \\ &=&4\int_{0}^{\pi /6}t\,dt+4\int_{0}^{\pi /3}\arctan \left( \tan \frac{\theta}{2} \right) d\theta \\ &=&\frac{\pi ^{2}}{18}+\frac{\pi ^{2}}{9}=\frac{\pi ^{2}}{6}. \end{eqnarray*} $
One of the substitutions is a new one. After the first pair of substitutions $x,y$, we have done two additional ones in the resulting integrals, as indicated above: in the 1st, $u=\sqrt{2}\sin t,du=\sqrt{2}\cos t\,dt$, and in the 2nd, $u=\sqrt{2}\cos \theta,du=-\sqrt{2}\sin \theta\,d\theta$.