The claim makes most sense if it takes place within a set theory that satisfies ZF, without any axiom of Regularity (so asking whether a set is well-founded is interesting), and with a form of Extensionality that allows urelements (so asking whether a set is pure is interesting).
The "only if" direction is easy enough -- you can prove by induction on $\alpha$ that all elements in $V_\alpha$ will be pure and well-founded.
For the "if" direction, first prove as a lemma that if $A$ is a set with $A\subseteq \mathbf V$, then $A\in\mathbf V$. (This uses the Axiom of Replacement to find a high enough level in the hierarchy to contain $X$).
Now assume that $X$ is pure and well-founded. Let $Y$ be the transitive closure of $X$ and assume for a contradiction that $Y\not\subseteq \mathbf V$. Then, since $\in$ is well-founded on $Y$, there will be a $\in$-minimal element of $Z$ of $Y$ that is not in $\mathbf V$. Since $Z$ is a set (all elements of $Y$ are sets rather then urelements because $X$ is pure), this means that $Z\subseteq V$, so by the lemma $Z\in V$, a contradiction.
Thus $Y\subseteq \mathbf V$, so $X\subseteq \mathbf V$, so $X\in \mathbf V$.