As Asaf and Joel have observed, the answer to your question is negative. However, there is a sense in which being an elementary submodel of $L_{\omega_1}$ is the only way to "persistently" get elementary submodelhood relations.

Specifically, the following are equivalent:

$L_\alpha\prec L_{\omega_1}$.

There is a club $S\subseteq\omega_1$ such that $L_\alpha\prec L_\beta$ for all $\beta\in S$.

But on the *other* other hand, if $V=L$ then there **is** an unbounded $U\subseteq \omega_1$ such that for all $\alpha,\beta\in U$ we have $L_\alpha\equiv L_\beta\not\equiv L_{\omega_1}$ *(note that this can't be proved using just a counting argument or forcing + absoluteness)*. This is a beautiful short application of Tarski's undefinability theorem due to Hjorth, answering question 10.4 of A. Miller. Hjorth's argument, with minor formatting edits from me, is copied below *(which I've left hidden to avoid spoilers)*:

Let $X$ be the set of complete theories that satisfy "everything is countable" and have unboundedly many $\alpha<\omega_1^L$ with $L_\alpha$ realising them. The theory of $L_{\omega_1^L}$ is one such theory, and we will be done if we prove that there are some others. Now $X$ is a definable class in $L_{\omega_1^L}$, and so it must have some other elements or else $L_{\omega_1^L}$ would admit a truth definition ($\varphi$ is true in $L_{\omega_1^L}$ iff the unique element of $X$ contains $\varphi$).