= Solution
Let $S$ consist of all <universal-existential sentences> entailed by $T$, and take $A_0\models S$. By (c), there is an <existential embedding> of $A_0$ into $B_0\models T$. By (b), $B_0$ embeds in $A_1$ with $A_0\preccurlyeq A_1$. After replacing structures by isomorphic copies, arrange these maps as inclusions. Since $A_1\equiv A_0$, it also satisfies $S$ and the construction can be repeated. This produces
$$
A_0\subseteq B_0\subseteq A_1\subseteq B_1\subseteq A_2\subseteq\cdots,
\qquad A_i\preccurlyeq A_{i+1},\quad B_i\models T.
$$
The two interleaved chains have the same union $D$. The <elementary chain theorem> gives $A_0\preccurlyeq D$. The $B_i$ are an ordinary chain of models of $T$, so the assumed closure under unions of chains gives $D\models T$. Therefore $A_0\models T$. This proves
$$
\boxed{\operatorname{Mod}(T)=\operatorname{Mod}(S),\qquad S\subseteq\forall_2.}
$$
Conversely, universal-existential axioms are preserved in such unions: a finite universal tuple occurs in one stage, which already supplies its existential witnesses. The quantifier-free matrix has the same truth value in that stage and in the union. This also explains the <inductive first-order theory> characterization.
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