How should I solve this integral with changing parameters?

Start by making a drawing of your domain. You can see that the $v$ is along the diagonal in the first quadrant, and $u$ is along the diagonal in the second quadrant. You can also see that the line between $(0,2)$ and $(2,0)$ is parallel to $u$, and intersects $v$ axis at $v=1$. So $v$ varies between $0$ and $1$ and $u$ varies between $-v$ and $v$.

Your domain is $$D=\{(x,y):0<x<2\,,\,0<y<2\,,\,x+y<2\}$$

Using change of variables $(x,y)\to(u,v)$ with $$u=\frac{x-y}{x+y}\,,\,v=x+y$$

The region is now $$R=\{(u,v):-1<u<1\,,0<v<2\}$$

Therefore,

\begin{align} \iint_D \exp\left({-\frac{x-y}{x+y}}\right)\,\mathrm{d}x\,\mathrm{d}y&=\frac{1}{2}\iint_R ve^{-u}\,\mathrm{d}u\,\mathrm{d}v \\\\&=\frac{1}{2}\int_0^2v\,\mathrm{d}v\int_{-1}^1 e^{-u}\,\mathrm{d}u \end{align}