In this question we assume that all formulae are in the language of $\sf ZFC$ and that $\sf ZFC$ is consistent.
Recall that we say that a formula $\varphi(x,y)$ represents a set-like class relation iff for every $x$ the class of all $y$ such that $\varphi(x,y)$ forms a set (for example, $y\in x,\,y\subset x,\,x=y,\,y=\varnothing$ are set-like, but $x\in y,\,x\ne y,\,x=x$ are not).
Consider the following proposition schema (we may call it the Schema of Transitive Closure for set-like relations):
For every set-like relation $\varphi$ and every set $u$ there is a set $v$ that is a superset of $u$ and is closed under $\varphi$. More formally, if $\varphi$ does not have any of the variables $z, u, v$ free, $$\forall x\exists z\forall y\left[\varphi(x,y)\Rightarrow y\in z\right]\,\Rightarrow\,\forall u\exists v\left[u\subseteq v\land\forall x\forall y\left(x\in v\land\varphi(x,y)\Rightarrow y\in v\right)\right]$$
If we add this schema as an axiom and drop the usual axioms of Pair, Union, Infinity and Replacement, can we prove the dropped axioms as theorems in this theory? Is this theory equivalent to $\sf ZFC$?