A sentence is called existential if it is of the form $\exists x_1 \ldots \exists x_n \ \phi(x_1, \ldots, x_n)$, where $\phi$ is quantifier free.
We know that (see Chang-Keisler "Model Theory", Proposition 5.2.2)
Theorem. If every existential sentence that holds in $\mathfrak{A}$ also holds in $\mathfrak{B}$, then $\mathfrak{A}$ is isomorphically embeddable in some elementary extension of $\mathfrak{B}$.
Suppose that we have a theory $T$ such that all its models agree on existential sentences. By the previous theorem, starting with two models $\mathfrak{A}, \mathfrak{B} \models T$, we can prove that for any $n > 0$, $\mathfrak{A}$ is $n$-sandwiched by model $\mathfrak{B}$. To see this, notice that since $\mathfrak{B} \models T$ and all models of $T$ agree on existential sentences, we have that $\mathfrak{B}$ can be isomorphically embedded in some elementary extension $\mathfrak{A}_0$ of $\mathfrak{A}$. This means that $\mathfrak{B} \subseteq \mathfrak{A}_0$. Now $\mathfrak{A}_0 \models T$, since $\mathfrak{A} \models T$ and $\mathfrak{A} \prec \mathfrak{A}_0$. So $\mathfrak{A}_0$ can be isomorphically embedded in some elementary extension $\mathfrak{B}_1$ of $\mathfrak{B}$. So we have $\mathfrak{B} \subseteq \mathfrak{A}_0 \subseteq \mathfrak{B}_1$. We can go on like this for infinitely many steps.
Is this argument wrong? If not, does this mean that every theory $T$ such that all its models agree on existential sentences, is also $\Pi_2$ (and actually even $\Pi_{2n}$) complete? Doesn't ZFC satisfy the above condition as $T$?
Note : Theorem (added). (Change-Keisler Proposition 5.2.5) If there are models $\mathfrak{A}', \mathfrak{B}'$ such that $\mathfrak{B} \subseteq \mathfrak{A}' \subseteq \mathfrak{B}'$, $\mathfrak{B} \prec \mathfrak{B}'$ and $\mathfrak{A} \prec \mathfrak{A}'$, then every $\Pi_2$ sentence holding in $\mathfrak{A}$ also holds in $\mathfrak{B}$.
I see a problem with your argument. When you say that $\mathfrak A_0$ can be embedded in an elementary extension of $\mathfrak B$, it means that there is an elementary extension $\mathfrak B_1$ of $\mathfrak B$ and an embedding $f : \mathfrak A_0 \to \mathfrak B_1$. Then you say that we can identify $\mathfrak A_0$ with $f(\mathfrak A_0)$ and get $\mathfrak B \subseteq \mathfrak A_0 \subseteq \mathfrak B_1$. The problem is that you don't know what $f$ does to $\mathfrak B$. In other words $f(\mathfrak B)$ might not be an elementary substructure of $\mathfrak B_1$.