= Solution
Write $S_L=X^{\oplus L}$. Its <Poisson distribution> has mean $L\lambda$, since the probability-generating functions of the independent summands multiply. Define
$$
Q_L=\frac{S_L-L\lambda}{L^{(1+\beta)/2}},\qquad a_L=L^\beta.
$$
The scaled <cumulant-generating function> is
$$
\Lambda_L(\theta)=\frac1{L^\beta}\log\mathbb E e^{L^\beta\theta Q_L}
=\lambda L^{1-\beta}\left(e^{\theta L^{(\beta-1)/2}}-1-\theta L^{(\beta-1)/2}\right).
$$
A <Taylor expansion> of the exponential, valid for each fixed $\theta$ because $\beta<1$, gives
$$
\Lambda_L(\theta)\longrightarrow\Lambda(\theta)=\frac\lambda2\theta^2.
$$
The remainder is $O(L^{(\beta-1)/2})$. The limit is finite and differentiable on all of $\mathbb R$, hence satisfies the essential-smoothness hypotheses of the <Gärtner–Ellis theorem>. Its <Legendre-Fenchel transform> is
$$
\boxed{I_\beta(x)=\sup_\theta\{\theta x-\lambda\theta^2/2\}=\frac{x^2}{2\lambda}.}
$$
This proves the <Poisson moderate deviation principle> with <large-deviation speed> $L^\beta$ and a <good rate function>.
For an arbitrary <open set> $B$, the lower bound uses $\inf_BI_\beta$ and the upper bound uses $\inf_{\overline B}I_\beta$. These infima agree: every point of the closure is a limit of points of $B$ and $I_\beta$ is finite and continuous. Thus the <logarithmic probabilities of open sets for a continuous rate function> give the stated logarithmic limit for every open $B$, not just intervals. The empty-set case follows from the usual infinite-value conventions.
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