If $R \in L_\alpha$ is a binary relation so that $L_\alpha$ thinks $R$ is well-founded, must $R$ truly be well-founded? (Edit) That is, if $L_\alpha$ thinks that every nonempty subset of the domain of $R$ has a least element, is the same true in $V$?

(Edit) If $L_\alpha$ satisfies a sufficiently strong fragment of $\mathsf{ZFC}$, then the answer is yes because then $L_\alpha$ can build a rank function from $R$ into its ordinals and thus an ill-foundedness in $R$ would give rise to an ill-foundedness in $\mathrm{Ord}$, which is impossible. But if $L_\alpha$ does not have rank functions for all well-founded relations then it only has access to the $\Pi_1$ characterization of well-foundedness. It is conceivable in this case that $L_\alpha$ could be wrong about well-foundedness, that it has a relation which it wrongly believes to be well-founded.

For comparison, Zermelo set theory (= $\mathsf{ZFC}$ minus Replacement and Foundation) has transitive models which are wrong about well-foundedness. The problem with these models is that they fail to have enough ordinals to capture the ordertype of every well-order.

That these bad models exist is an easy consequence of a theorem by Harvey Friedman.

Theorem(H. Friedman, 1973): Fix a countable admissible set $A$. Consider $T$, a theory extending $\mathsf{KP}$ in the infinitary logic $L_A$ which has a model containing $A$ and is $\Sigma_1$-definable over $A$. Then there is an ill-founded $M \models T$ so that the ordinals of the well-founded part of $M$ are exactly the ordinals of $A$ and the well-founded part of $M$ contains $A$.

To build these bad models of Zermelo set theory, consider the theory consisting of $\mathsf{KP}$ plus the assertion that $V_{\omega + \omega}$ exists. This theory has models containing your favorite countable admissible set, which let's say is $L_{\omega_1^{CK}}$. Let $M$ be the model Friedman's theorem produces when applied to this theory and admissible set and consider $N = V_{\omega+\omega}^M$. Then $N$ is transitive, since everything in $N$ has rank far less than $\omega_1^{CK}$, and $N$ agrees with $M$ about well-foundedness. Take any countable "ordinal" from the ill-founded part of $M$ and you can find an isomorphic copy, call it $R$, which is a subset of $\omega^2$ and hence in $N$. Then, $N$ wrongly thinks $R$ is well-founded.

This argument won't answer the question for non-admissible $L_\alpha$s. Any well-order in $L_\alpha$ must have ordertype less than the least admissible $\beta > \alpha$ as otherwise $L_\beta$ would see an isomorphic copy of its ordinals in an initial segment of itself. If we tried to run a variation of the above argument to produce an ill-founded $M$ with $L_\alpha$ in its well-founded part, we would have that the well-orders in $L_\alpha$ have ordertype in the well-founded part of $M$. As such, we cannot by this means produce an $L_\alpha$ which is wrong about well-foundedness.

(Edit) As pointed out by François and Noah, admissibility isn't sufficient to make $L_\alpha$ correct about well-foundedness. The particular case I'm interested in is when $\alpha$ is the successor of an ordinal whose corresponding fragment of $L$ is correct about well-foundedness.

Question: (Edit) Is there $\xi$ so that $L_\xi$ satisfies enough of $\mathsf{ZFC}$ to be correct about well-foundedness but $L_{\xi+1}$ is wrong about well-foundedness?