This statement has complexity $\Delta_2$, because it is a locally verifiable feature, meaning one that can be decided yes-or-no inside any sufficiently large $V_\theta$. Any $V_\theta$ above $\alpha$ and $\beta$ will correctly determine whether $V_\alpha\equiv V_\beta$.

I wrote a blog post about local properties in set theory, which explains further details about the complexity calculation.

Let me add that most of the complexity of $V_\alpha\equiv V_\beta$, however, as a relation on $\alpha$ and $\beta$, is just knowing that you have the right $V_\alpha$ and $V_\beta$. In contrast, the relation $M\equiv N$, as a relation on $M$ and $N$, is $\Delta_1$, since you just have to say that the unique satisfaction relations for these structures fulfill the same sentences. Because the satisfaction relations are unique in fulfilling a certain $\Delta_0$ expressible recursion, we can express this in a $\Sigma_1$ way or a $\Pi_1$ way, making them $\Delta_1$. Consequently, your $\Pi_1$ case would be fine if we view it as a relation on pairs $(V_\alpha,V_\beta)$, rather than as a relation on $(\alpha,\beta)$.

One can see that $V_\alpha\equiv V_\beta$ is not complexity $\Pi_1$ in $(\alpha,\beta)$ in all models of set theory by observing that it is not always downwards absolute. For example, perform an Easton-support product to form an extension $V[G]$ by forcing a violation of GCH at every regular cardinal. Now by cardinality considerations there will be ordinals $\alpha\neq \beta$ such that $V[G]_\alpha$ and $V[G]_\beta$ have the same theory, where $\alpha=\kappa+3$ for some regular cardinal $\kappa$. If $V[G^-]\subseteq V[G]$ is the submodel that omits the forcing at coordinate $\kappa$, then $V[G^-]_\alpha$ is not elementarily equivalent to $V[G^-]_\beta$, since the former knows that the missing coordinate is very near the top, but the latter doesn't.