I sketch the arguments for $C(x)$, the arguments for $L(x)$ are essentially the same.
The specific form of the sum suggests probabilistic arguments. Let $X_x$ be a $\mathrm{Poiss}(x^2)$-distributed random variable and note that $$C(x)\,e^{-x^2}=\mathbb{E}\frac{x}{\sqrt{X_x}}\,1_{\{X_x\geq 1\}}$$ It is known that $\frac{X_x -x^2}{x}\longrightarrow N(0,1)$ (standard normal) in distribution as $x\longrightarrow \infty$ and that that all moments $\mathbb{E}\left(\frac{X_x-x^2}{x}\right)^k$ of $X_x$ converge to the corresponding moments of $N(0,1)$, so that (for large $x$) $X_x$ is concentrated around $x^2$, with deviations of order $x$.
To use that information split $\{X_x\geq 1\}$ on the rhs into (say) the parts $1\leq X_x <\tfrac{1}{2}x^2$, $X_x-x^2 >\tfrac{1}{2} x^{2}$ and $|X_x-x^2| \le \tfrac{1}{2}x^{2}$.
By routine arguments the integrals over the first two parts are asymptotically exponentially small. For the remaining part write \begin{align*} \mathbb{E}\frac{x}{\sqrt{X_x}}\,1_{\{|X_x-x^2|\leq \tfrac{1}{2}x^2\}} &=\mathbb{E}\frac{x}{\sqrt{x^2+(X_x-x^2)}}\,1_{\{|X_x-x^2|\leq \tfrac{1}{2} x^2\}}\\ &=\mathbb{E}\frac{1}{\sqrt{1 +\frac{1}{x^2}(X_x-x^2)}}\,1_{\{|X_x-x^2|\leq \tfrac{1}{2} x^2\}}\\ &=\mathbb{E}\sum_{k=0}^\infty {-\frac{1}{2} \choose k} \frac{1}{x^{2k}}(X_x-x^2)^k \,1_{\{|X_x-x^2|\leq \tfrac{1}{2}x^2\}}\\ %C(x)\,e^{-x^2}&= 1+\frac{3}{8}x^{-2} + \frac{65}{128}x^{-4} + \frac{1225}{1024}x^{-6} + \frac{1619583}{425984}x^{-8}+\mathcal{O}(x^{-10})\\ %L(x)\,x^2\,e^{-x^2}&= 1+\frac{15}{8}x^{-2} + \frac{665}{128}x^{-4}+\frac{19845}{1024}x^{-6}+\frac{37475823}{425984}x^{-8}+\mathcal{O}(x^{-10}) \end{align*} Clearly the series may be integrated termwise, and completing the tails changes it only by asymptotically exponentially small terms. For $k\geq 2$ the central moment $c_k(x):=\mathbb{E}\left(X_x-x^2\right)^k$ is a polynomial in $x^2$ of degree $\lfloor k/2\rfloor$. Thus (after regrouping of terms) the formal series $$\sum_{k=0}^\infty x^{-2k}{-\tfrac{1}{2} \choose k} c_k(x)$$ gives a full asymptotic expansion of $C(x)e^{-x^2}$. Evaluating the first eight terms gives $$C(x)\,e^{-x^2}= 1+\frac{3}{8}x^{-2} + \frac{65}{128}x^{-4} + \frac{1225}{1024}x^{-6} + \frac{131691}{32768}x^{-8}+\mathcal{O}(x^{-10})$$
EDIT: I corrected the coefficient of $x^{-8}$. Thanks to Johannes Trost for pointing out that I miscalculated. Similarly the coefficient of $x^{-8}$ in the asymptotic series for $L(x)$ given in the comment must be corrected.