I think the "correct" answer is that the question is misguided, but I'm going to try to give lots of alternate answers in case one of them makes you happier.
In ZFC set theory as usually phrased, there are no terms at all in the sense of logic, other than variables (i.e. there are no function symbols in the logical signature). There are only axioms which assert that sets satisfying certain properties exist (and are unique). For instance, any expression involving $\bigcup x$ is shorthand for a statement about any (hence the unique) set which contains exactly those $y$ such that $y\in z\in x$ for some $z$. That's even true about the empty set symbol $\emptyset$! So any statement about "$TC(x)$" will also have to formally be a statement about any (hence the unique) set which is a transitive closure of $x$.
One might try to make all of the axioms of ZFC into term-forming operators, so that instead of saying "there exists a set with no elements" there would be a specified term $\emptyset$ and an axiom saying "$\emptyset$ has no elements," and likewise for pairings, unions, replacement, etc. (Of course the operator for replacement will have to take a first-order formula as its input as well as a set.) In that case you should be able to apply the "replacement operator" followed by the "union operator" in order to construct a term describing $x \cup \bigcup x \cup \bigcup\bigcup x \cup \dots$.
Alternately, one can add a choice operator such that for any formula $\phi$, the term $\varepsilon x. \phi(x)$ has the property that if there exists an $x$ with $\phi(x)$, then in fact $\phi(\varepsilon x.\phi(x))$. Then you can define $TC(x) = \varepsilon y. ISTC(y,x)$. In this case, one could even restrict to a unique choice operator which only applies to formulas $\phi$ such that there is at most one $x$ satisfying $\phi(x)$.
We can also write $TC(x)$ in the undergraduate's "set-builder notation:"
$$ TC(x) = \Big\lbrace y \;\Big\vert\; (\forall z)\; \Big( (\forall a,b)\; a\in b \wedge b\in z \Rightarrow a\in z\Big) \wedge x \subseteq z \Longrightarrow y\in z \Big\rbrace $$
but of course ZFC does not include general set-builder notation as a term-forming operation, nor can it be extended to do so, since not every set-builder notation forms a set (e.g. $\lbrace x \mid x\notin x\rbrace$).