Suppose we started with a set $A$ and wanted to define the closure T of $A$ under Replacement. Does the following recursive definition work?
$T_0 = A$
$T_{\alpha+1} = T_{\alpha} \cup\hspace{2mm} \{t\hspace{1.5mm} |\hspace{2mm} \exists x \in X (f(x)=t) : X \in T_{\alpha} ,f:X \rightarrow V \}$
$T_{\alpha} = \bigcup_{\beta<\alpha} T_\beta$ for $\alpha$ limit.
Let $T = \bigcup_{\alpha \in \textrm{ORD}} T_{\alpha}$
Then $T \models$ Replacement.
Proof: $T$ is transitive by construction, hence it suffices to show that whenever $X \in$ T and $f: X \rightarrow T$, then $rng(f) \in T$. So, let $X \in$ T and $f: X \rightarrow T$. Then $X \in T_{\alpha}$ for some $\alpha$. Let $v \in rng(f)$. Then $\exists u$ such that $f(u) = v$. Since $dom(f) = X$ and $X \in T_{\alpha}$, $u \in X$. Hence $v \in T_{\alpha+1}$. Therefore $rng(f) \subseteq T_{\alpha+1} \in$ T. So $rng(f) \in T$ as required.