$f_{X_1} = \exp(-x_1)$, $f_{X_2} = \exp(-x_2)$, $X_1$ and $X_2$ are independent.
$Y_1 = X_1 - X_2$ and $Y_2 = X_1 + X_2$
$0 < x_i <\infty$
This gives me the joint $f_{Y_1Y_2} = \frac{1}{2} \exp(-y_2)$.
Using the supports I obtain $y_1<y_2<\infty$, and $0<y_2<\infty$. However using this to obtain the marginal $f_{Y_2} = \int^{y_2}_0 f_{Y_1Y_2} dy_1 = \frac{1}{2}y_2 \exp(-y_2)$ shows I must be mistaken as the integral over the support $\neq$ 1
Your joint density $$f_{Y_1,Y_2}(y_1,y_2)= \frac{1}{2} \exp(-y_2)$$ applies when $0 \le y_2$ and $-y_2 \le y_1 \le y_2$, and is $0$ otherwise. So the marginal distribution is $$f_{Y_2}(y_2)=\int\limits _{y_1=-y_2}^{y_2}f_{Y_1,Y_2}(y_1,y_2) \, dy_1 =\int\limits _{y_1=-y_2}^{y_2} \frac{1}{2} \exp(-y_2) \, dy_1 = y_2\exp(-y_2)$$ when $0 \le y_2$ and is $0$ otherwise.
This is the density of a Gamma distribution with shape parameter $2$ and does integrate to $1$