So my proof from a textbook reads:
"$\operatorname{Aut}(G)$ is a subset of $S_G$ and since $id \in \operatorname{Aut}(G)$, we may apply a subgroup test [Let G be a group and H be a nonempty subset of G. If $\forall a,b\in H, ab^{-1}\in H$, then H is a subgroup of G].
Suppose $\alpha, \beta \in \operatorname{Aut}(G)$. Then $\alpha\beta: G\to G$ and $\alpha^{-1}$ are elements of $S_G$, so they are bijective. Therefore it is sufficient to show the operation preserving property."
Why is it enough to show the operation preserving property?
It means that all one needs to do is show that $\alpha\beta$ and $\alpha^{-1}$ are homomorphisms, since it is already the case that they are bijective on $G$.
It also suffices to prove that $\alpha\beta^{-1}$ is a homomorpism, by the one-step subgroup test.