We already know that the models of $\sf ZFC_2$ are exactly the sets $V_\kappa$ for $\kappa$ inaccessible, so we really just want an axiom $\varphi$ such that $\sf ZFC_2+\varphi$ would only have one inaccessible cardinal which satisfies it.
For example $\varphi$ could be "There are no inaccessible cardinals" (satisfied by the first inaccessible); or "There is exactly one inaccessible cardinal" (satisfied by the second inaccessible); and so on.
Namely if $K$ is the set of inaccessible cardinals then $\varphi$ should tell us what is $K\cap V_\kappa$ in a unique way. This is not necessarily possible when that set is complicated enough (or when $\kappa$ is large enough that all sort of crazy reflections are going on), but if we can characterize $K\cap V_\kappa$ uniquely then the axiom guarantees that at most one $V_\kappa$ can satisfy that axiom.
Some words on the issue of $\sf GCH$. Suppose that the universe of set theory $V$ satisfies $\sf GCH$, in this case it is clear that any $V_\kappa$ satisfies $\sf GCH$ as well, and in particular for inaccessible $\kappa$, which means that every model of $\sf ZFC_2$ is a model of $\sf GCH$.
If, on the other hand, $V$ does not satisfy $\sf GCH$ and the example for the failure is below the first inaccessible cardinal of $V$ then for sufficiently big $\kappa$ (which is still much smaller than the inaccessible cardinals) $V_\kappa$ fails to satisfy $\sf GCH$. In particular all the models of $\sf ZFC_2$ would.
Think of something worse, suppose there are exactly two inaccessible cardinals, $\kappa_1<\kappa_2$. $\sf GCH$ holds below $\kappa_1$ but fails at $\kappa_1$. It follows that $V_{\kappa_1}\models\sf ZFC_2+GCH$ and $V_{\kappa_2}\models\sf ZFC_2+\lnot GCH$. And $V_{\kappa_1}$ is the unique model of $\sf ZFC_2$ in which there are no inaccessible cardinals, whereas $V_{\kappa_2}$ is the unique model of $\sf ZFC_2$ in which there is exactly one inaccessible cardinal.
The reason for this is that full semantics second-order logic does not have the completeness theorem. The fact there is just one model up to isomorphism does not mean that the theory can prove or disprove any proposition in the language.