Is the following proof sound/does anyone have another more elegant (categorical) proof?
The direction $\Rightarrow$ is obvious the "family of stalks"-functor is a functor and functors take isos to isos.
$\Leftarrow$: let us first show we get $\mathcal F \rightarrow \mathcal G$ is mono. We have the diagram
$\require{AMScd}$\begin{CD} \mathcal F(X) @>{\phi}>> \mathcal G(X)\\ @VVV @V{\mu}VV \\ \prod \mathcal F_p @>\bar\phi>> \prod \mathcal G_p \end{CD}
where the vertical and the bottom maps are mono. It follows that $\mu\phi$ is mono; hence $\phi$ is mono.
Now we show that $\phi$ is epi. An element in $\mathcal G(X)$ consists of a colection of compatible germs in $\prod \mathcal G_p$. Since the map on the bottom is an isomorphism, we have a corresponding collection in $\prod\mathcal F_p$; if it is also compatible it will lift up to $\mathcal F$. But it is compatible because it the isomorphism is induced by a morphism upstairs. More explicitly:
Take two germ representatives $(U^q,g^q), (U^t,g^t)$ of a compatible collection in $\prod \mathcal G_p$ such that their preimages via $\bar\phi$ are germs defined over the same open sets $(U^q,f^q), (U^t,f^q)$. We want to show that $$f^q - f^t = 0\qquad \text{in}\, U^q\cap U^t.$$ Since $\bar\phi$ is induced by a morphism of sheaves $\phi$, we get
$$\bar\phi:\prod_{p\in U^q \cap U^t} (f^q-f^t)_p\mapsto\prod_{p\in U^q \cap U^t}(g^q-g^t)_p$$
by looking at appropriate representatives $(V_p,\text{res}$$\vert_{Vp}(f^q-f^t))$ of $(f_q-f_t)_p$, where $V_p\subset U_q \cap U_t$ is small enough that one would be able to find a representative of the germ $(g^q-g^t)_p$ over it. By assumption, the right-hand side is zero and since $\bar\phi$ is an iso so is the left-hand side. We have what we wanted.