 Solution

Treat $0$ as the trivial group if $G$ is not abelian. The restriction maps satisfy the presheaf identities. To verify the sheaf gluing axiom, take an open cover $(V_i)$ of $U$. If $x\notin U$, every section group is trivial. If $x\in U$, at least one $V_i$ contains $x$. Every two such opens have an overlap containing $x$, so compatibility forces all their sections to have the same value $g\in G$. Opens not containing $x$ have their unique trivial section. The value $g\in \mathcal{G}(U)$ is exactly the unique glued section. Thus the stated presheaf is a skyscraper sheaf, including for nonabelian groups.

For the stalk of a sheaf, take the direct limit over neighbourhoods of $y$. If some neighbourhood $V$ of $y$ omits $x$, neighbourhoods inside $V$ are cofinal and all their groups are trivial. If every neighbourhood of $y$ contains $x$, every group in the system is $G$ and every transition map is the identity. Consequently, on the arbitrary topological space in the question,Membership in this closure means exactly that every neighbourhood of $y$ contains $x$. In particular ${\mathcal{G}}_x=G$. If $x$ is a closed point, the closure is just $\{x\}$ and the familiar point-supported answer results. The separation assumption is not present in the printed question, so it cannot be imposed silently.