The answer is yes. Indeed, for any Schwartz function $\phi$ we have $(h*G)(\phi)=G(h^-*\phi)=G(h*\phi)$, where $h^-(x,y):=h(-x,-y)=h(x,y)=e^{-x^2-y^2}$. So,
\begin{multline}
(h*G)(\phi)=G(h*\phi)=\int dx\,f(x)g(x)\iint du\,dv\,\phi(u,v)e^{-(x-u)^2-v^2} \\
-\int dy\,f(y)g(y)\iint du\,dv\,\phi(u,v)e^{-u^2-(y-v)^2} \\
=\int dx\,f(x)g(x)\iint du\,dv\,\phi(u,v)[e^{-(x-u)^2-v^2}-e^{-u^2-(x-v)^2}] \\
=\iint du\,dv\,\phi(u,v)p(u,v),
\end{multline}
where
\begin{equation}
p(u,v):=\int dx\,f(x)g(x)[e^{-(x-u)^2-v^2}-e^{-u^2-(x-v)^2}].
\end{equation}
We have
\begin{multline}
\iint du\,dv\,|p(u,v)|\le2\int dx\,|f(x)g(x)|\iint du\,dv\,e^{-(x-u)^2-v^2} \\
=2\int dx\,|f(x)g(x)|\,c^2<\infty,
\end{multline}
where $c:=\int dv\,e^{-v^2}<\infty$.
So,
\begin{equation}
(h*G)(\phi)=\iint p\phi,
\end{equation}
and $p\in L^1$.