TITLE: Is this set theory equivalent to ZFC?
QUESTION [11 upvotes]: Consider a variant of set theory with these axioms:

Extensionality,
Regularity (foundation),
Separation,
Powerset,
Axiom of Choice, and
Transitive closure of a set-like relation is set-like. Update: This did not exactly represent what I had in mind, so the corrected version is given on the next line, and its precise formalization is given below. Sorry for my mistake and ensuing confusion.
The transitive closure of any set under a set-like relation is a set.

Note that it does not explicitly postulate Pairing, Union, Infinity and Replacement.
Question: Is this set theory equivalent to $\mathrm{ZFC}$?

Detailed explanation and formalization:

We use symbol $\prec$ to represent a binary relation. In general, it is a definable class relation, that is a first-order formula with 2 free variables (and, possibly, additional parameters). As usual, we write $a\prec b$ to represent $\prec\!(a,b),$ and we assume that all bound variables in any formula are automatically renamed before a substitution to avoid variable name conflicts that would change its meaning.
We use “is a set” and “exists” as synonyms; “sethood” and “existence” are also synonyms.
We write $a\prec b\prec c$ to represent $a\prec b\land b\prec c$. This notation can also mix several different relation symbols, e.g. $a\prec b\in c$.
When we say that a relation $\prec$ is “set-like”, we mean
$$\color{green}{\forall x\,\exists y\,\forall z\left(z\prec x\;\Rightarrow\; z\in y\right)}.$$
When we say that “$w$ is a superset of the transitive closure of $s$ under the relation $\prec$”, we mean
$$\color{maroon}{s\subseteq w\,\land\,\forall u\,\forall v\left(u\prec v\in w\;\Rightarrow\; u\in w\right)}.$$
We also may rephrase it as “the transitive closure of $s$ under $\prec$ is a subset of $w$” or simply “the transitive closure of $s$ under $\prec$ is a set”. At this point, we do not need to define what “the transitive closure” exactly is, because we are only interested in asserting its sethood, so existence of any its superset $w$ is sufficient for our purposes. I suppose that, when the need arises, “the transitive closure” can be defined as the smallest such set, and can be carved out of its superset using Separation.
Our last axiom asserts that, provided $\prec$ is a set-like relation, the transitive closure of any set $s$ under that relation $\prec$ is a set. It can be formalized using the following axiom schema where $\prec$ ranges over all binary relations:
$$\left(\vphantom{\Large|}\color{green}{\forall x\,\exists y\,\forall z\left(z\prec x\,\Rightarrow\,z\in y\right)}\right)\,\Rightarrow\,\forall s\,\exists w\!\left(\vphantom{\Large|}\color{maroon}{s\subseteq w\,\land\,\forall u\,\forall v\left(u\prec v\in w\,\Rightarrow\,u\in w\right)}\right)\!.$$

REPLY [12 votes]: Accepting the convention that it is a logical axiom that the universe is nonempty, the answer is yes. We will formalize the transitive closure axiom schema (TC) as follows: for any definable (with parameters) binary relation $R,$ if for all $x,$ $\{y: y R x\}$ is a set, then for all $x,$ there is a set $T$ such that $x \in T$ and $T$ is closed downwards under $R.$ (*) Of course, this can only be weaker than asserting the existence of a minimum such $T.$
For efficiency, we will prove Pairing, Union, Infinity, and Replacement from Extensionality, Separation, and TC.
Pairing: We first note that $\emptyset$ exists by applying separation to an arbitrary set. Next, for all $x,$ $\{x\}$ exists by applying TC to $x$ and the empty relation. Finally, for all $x, y,$ we get $\{x,y\}$ by applying TC to $x$ and the relation defined by $a R b$ iff $b = x$ and $a=y.$
Union: Fix a set $S.$ By Separation and Russell's paradox, there is $x \not \in S.$ Define $R$ by $a R b$ iff $b = x$ and $a \in S$ or $b \in S$ and $a \in b.$ Then we get $\bigcup S$ by applying TC to $x$ and $R.$
Infinity: Define a relation $R$ by $a R b$ iff $a$ and $b$ are natural numbers and $a=b+1.$ Then $\omega$ exists by applying TC to $\emptyset$ and $R.$
Replacement: Fix a set $S$ and a definable function $F.$ Fix $x \not \in S.$ Define $R$ by $a R b$ iff $b=x$ and $a \in S$ or $b \in S$ and $a = F(b).$ Then we get $F"S$ by applying TC to $x$ and $R.$
(*) Note that my formulation of TC only makes sense under the convention that the transitive closure of a relation is reflexive. Without this convention, then it's not clear we can prove the existence of $\{x\}$ from the axioms I specified. Of course, we can prove it exists from Separation and Power Set, which is included in the axioms listed in the question, but that feels overpowered for our purposes.
Edit: The question was updated with the intended formalization of the transitive closure schema. My TC here follows from Vladimir's version plus existence of $\{x\}$ for all $x,$ and the latter follows from Separation and Power Set.