(I'm a student so what i'm going to write next can hide any kind of errors...but i hope not. Moreover this is one of my first answers here, so tell me if it is someting wrong in my exposition or anything else. I'll be gratefull with anyone who show me errors in my argument. Thanks.)

I think you don't need $V=L$ to conclude $\forall x\in V\quad tc(x)\in V$ (where $tc(x)$ is the transitive closure of $x$).

As Kunen has proved in his "Set Theory" (III-Th.3.6), in $ZF^-$ (i.e. ZF without Foundation) you can prove that
$$x\in WF\equiv tc(x)\in WF,$$
where WF is the class of well founded sets. Then (by III-Th.4.1) you have that in $ZF^-$ you can prove that

$$\text{axiom of Foundationn}\equiv V=WF.$$

so if you are in $ZF$ you have Foundation and so you have $V=WF$ (i.e. $\forall\,x\quad x\in WF$) and so you have $\forall x\quad tc(x)\in V$ that is the same as $\forall\,x\; \exists\,y\quad y=tc(x)$. cvd

Note that, if you work in $ZF^-$, you mantain the result in the case of sets in $WF$ (i.e $\forall x\in WF\quad tc(x)\in WF(\subseteq V)$) but, in general you cannot say that for all $x\in V$. At this step if you assume $V=L$ than you can use VI-Lemma 1.11 (i.e $L(\alpha)\subseteq R(\alpha)$ for all $\alpha$) and so you can conclude that $L\subseteq WF (\subseteq V)$ and by VI-Lemma 1.6 (i.e. $L(\alpha)$ is transitive and $\forall\,\xi\leq\alpha\;\quad L(\xi)\subset L(\alpha)$) and the recursive costruction of $L$, you can conclude both $L$ is transitive and $\forall\,x\in L\quad ct(x)\in L$

Moreover the definition of $L$ is based on $Df$ (and/or $En$; see Def.V-1.1,1.4) and by Lemma V-1.7 thay are absolute for transitive model of "ZF without Power Axiom", and $WF$ is a model of whole ZF.

So, in my point of view, this "prove" that $L$ is closed under transitive closure (and so V is close too, if you assume L=V) with or without Foundaton, and you don't need to assume any kind of transitive closure of it.