There is a $\lambda<\omega_1$ such that $L_\lambda\prec L_{\omega_1}$, this can be obtained by taking the least ordinal not in $\textrm{Hull}^{L_{\omega_1}}(\{\varnothing\})$. According to [Wikipedia](https://en.wikipedia.org/wiki/Condensation_lemma) the hull is already transitive, so no Mostowski collapse is needed to put it in the form $L_\lambda$. Such $\lambda$ are unbounded in $\omega_1$: for any given $\xi<\omega_1$, by taking $\textrm{Hull}^{L_{\omega_1}}(\xi\cup\{\xi\})$ instead, a $\lambda_\xi>\xi$ can be produced such that $L_{\lambda_\xi}\prec L_{\omega_1}$.

Then, the ordinals $\alpha<\omega_1$ such that there is a $\beta<\alpha$ with $L_\beta\prec L_\alpha$ are unbounded in $\omega_1$. In particular, if $\beta,\alpha<\omega_1$ are such that $L_\beta\prec L_{\omega_1}$, $L_\alpha\prec L_{\omega_1}$, and $\beta<\alpha$, then $L_\beta\prec L_\alpha$. (This may be shown routinely by passing $\phi(\vec a)$ between $L_\beta$, $L_{\omega_1}$, and $L_\alpha$, with $\vec a$ a sequence of parameters from $L_\beta$.)

Since $\omega_1$ has cofinality $\omega_1$, this unbounded subset of $\omega_1$ must have order type $\omega_1$. If $\lambda<\omega_1$ is such that $L_\lambda\prec L_{\omega_1}$, then each such $\{\beta,\alpha\}$ from the second paragraph is a finite $Lan$-model with a monomorphism to $\{\lambda,\omega_1\}$, so the order type of $S$ is $\omega_1$.