= Solution
For the specified <change of variables>,
$$
\frac{\partial(x,y)}{\partial(u,v)}=\det\begin{pmatrix}(1+v)/2&u/2\\(1-v)/2&-u/2\end{pmatrix}=-\frac u2.
$$
Thus the inverse-coordinate <Jacobian determinant> is $\partial(u,v)/\partial(x,y)=-2/u$ where $u>0$, and the area element uses the absolute value $dx\,dy=(u/2)du\,dv$.
Now $u=x+y$, $v=(x-y)/(x+y)$ and $xy=u^2(1-v^2)/4$. The positive quadrant maps to the half-strip $u\geq0$, $-1\leq v\leq1$. The original origin corresponds to the collapsed boundary $u=0$, where $v$ is not unique; this measure-zero degeneracy does not affect the integral. Therefore
$$
I=\frac14\int_0^\infty u^2e^{-u}\,du\int_{-1}^1\sqrt{1-v^2}\,dv=\frac14(2!)\left(\frac\pi2\right)=\boxed{\frac\pi4}.
$$
The second factor is the area of a unit semicircle. Independently, the original integrand factors into two identical <gamma integrals>, giving $\Gamma(3/2)^2=\pi/4$.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2014-resit-sum-difference-integration-region.png]
{title=Positive-quadrant coordinates mapped to an unbounded half-strip}
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