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 $a = x$ and $b=y.$
Union: Fix a set $S.$ By Separation and Russell's paradox, there is $x \not \in S.$ Define $R$ by $a R b$ iff $a = x$ and $b = S$ or $a \in S$ and $b \in a.$ 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 $a=x$ and $b=S$ or $a \in S$ and $b = F(a).$ 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.