I think so (even in dimension higher than 2, assuming that $F$ and $G$ are still finite sets, and not codimension two subvarieties of course). Let $H$ be an ample divisor in $X$ avoiding $F$ and let $H'$ be its strict transform in $Y$. Then $H'$ is ample in $Y$. In particular $Y={\rm Proj}(Y,H')={\rm Proj}(X,H)=X$.