First off, a correction: you're conflation $L_\alpha$ with $L_\alpha\cap\mathcal{P}(\omega)$. E.g. $L_{\omega_1^{CK}}$ is not the set of hyperarithmetic reals, but $L_{\omega_1^{CK}}\cap\mathcal{P}(\omega)$ is.
Second, for simplicity let's assume $\mathsf{V=L}$. This means that $L_{\omega_1}$ contains every real, so the analytic hierarchy stops well before $L_{\omega_1}\cap\mathcal{P}(\omega)$.
OK, what about smaller admissible ordinals - say, the $\omega_n^{CK}$s for finite $n$ - and the first couple levels of the analytical hierarchy? Well, already in $L_{\omega_2^{CK}}$ we have Kleene's $\mathcal{O}$, and so $L_{\omega_2^{CK}}\cap\mathcal{P}(\omega)$ contains every $\Pi^1_1$ real and every $\Sigma^1_1$ real. But in fact those reals already showed up in $L_{\omega_1^{CK}+1}$, so $L_{\omega_2^{CK}}\cap\mathcal{P}(\omega)$ is much bigger than (the Boolean algebra generated by) the $\Sigma^1_1$ and $\Pi^1_1$ sets. The question is: does it reach as high as $\Delta^1_2$?
The answer, perhaps surprisingly, is no. The gulf between the first two levels of the analytical hierarchy is gigantic. The key tool here is the canonical well-ordering of $L$. For any first-order sentence $\sigma$, let $r_\sigma$ be the $\le_L$-least real coding a level of $L$ satisfying $\sigma$; this is a $\Delta^1_2$ real, and consequently all the "easily-describable" levels of the $L$-hierarchy will fall short of containing all the $\Delta^1_2$ reals. For example, let $\theta$ be the first recursively Mahlo ordinal; then letting $\sigma$ be the sentence such that the $L$-levels satisfying $\sigma$ are exactly those with recursively Mahlo index we get that $r_\sigma$ is a $\Delta^1_2$ real not in $L_\theta$ (this last bit uses the fact that if $r_\sigma$ were in $L_\theta$, then $L_\theta$ would contain a real coding $L_\theta$ itself, which via the pointwise-definability of $L_\theta$ would contradict Tarski's theorem).
Hinman's book Recursion-theoretic hierarchies contains a lot of useful information about the $\Delta^1_2$ reals and just how complicated they can be. In particular, I recommend taking a look at Corollary V.4.11, which in a precise sense says that the $\Delta^1_2$ sets can't be "built from below" in as simple a way as the hyperarithmetic sets. This old answer of mine is also relevant for getting a sense of just how powerful the second level of the analytical (also called projective) hierarchy is.
Best Answer
Yes, $L$ contains all the ordinals as elements. In fact this follows from what you already know (that $L$ contains all the ordinals as subsets): since smaller ordinals are elements of larger ordinals, for every class $X$ and ordinal $\alpha$ we have that if $\alpha+1\subseteq X$ then $\alpha\in X$. So the "ambiguity" around the notion of containment here, isn't.
A bit more explicitly, you can show by transfinite induction that $\alpha\in L_{\alpha+1}\setminus L_\alpha$ for each ordinal $\alpha$. So this tells us exactly when each ordinal appears in $L$. Note that this exactly matches the behavior of the $V$-hierarchy: in general $L_\alpha$ is always "as tall as" $V_\alpha$ but usually much "narrower" (even if $V=L$ - note that $L_\alpha$ is countable whenever $\alpha$ is countable, while $V_{\omega+1}$ already has size continuum).