This is another, totally different (and correct !) approach for answering the question. It is simply too long for a comment. So I decided to write it in a new answer. (Although that might look odd, but my previous answer, though accepted, is wrong.)
First define $$ I_{\mu}(y) = y^{1/2} \sum_{n=1}^{\infty} \ \frac{y^{n}}{n! \ n^{\mu}}, $$ and observe that the OP's functions are $$ C(x)= I_{\frac{1}{2}}(x^{2}) $$ and $$ L(x)=I_{\frac{3}{2}}(x^{2}) $$ Let $m$ be the integer part of $\mu$ and $\xi$ the fractional part, $0<\xi<1$.
Replace $n^{-\mu}$ in the definition of $I_{\mu}(y)$ by a ratio of $\Gamma$-functions times an asymptotic series in $n$ using this formula and for the coefficients of the asymptotic series in $n$ using the Norlund polynomials, $B_k^{(\alpha)}(x)$ found here. They are available in Mathematica via $\mathtt{NorlundB[k,\alpha,x]}$. Concretely, $$ n^{-\xi}\sim \frac{\Gamma(n)}{\Gamma(n+\xi)}\sum_{k=0}^{\infty}n^{-k} {\xi \choose k} B_k^{(1+\xi)}(\xi). $$
Now exchange the summation of the asymptotics in $n$ (with, say, summation index $k$) and the summation over $n$, which I assume light heartedly to be possible. The (now inner) sum over $n$ results in generalized hypergeometric functions of the form $$ _{m+k+2}F_{m+k+2}\left(\left. {1,\dots,1}\atop{2,\dots,2,1+\xi} \right\vert y\right), $$ with $m+k+2$ 1s in the upper line and $k+m+1$ 2s in the lower line. To get there one has to shift the summation index such that summation starts at $n=0$. Then insert $n+1=\frac{(2)_{n}}{(1)_{n}}$, with the usual Pochhammer symbols used. The defining formula for generalized hypergeometric functions results.
Using the asymptotic expansion of the generalized hypergeometric function for $y\rightarrow \infty$ given in a paper by Volkmer and Wood (downloadable from here) and after some simplifications one arrives at the asymptotic formula $$ I_{m+\xi}(y)=e^{y}\ y^{\frac{1}{2}-m-\xi}\ \frac{\xi\ \sin \pi\xi}{\pi}\sum_{k=0}^{\infty}(-1)^{k+1} y^{-k} \frac{\Gamma(k-\xi)}{k!}\ B_{k}^{(1+\xi)}(\xi) \\ \sum_{s=0}^{\infty} y^{-s} \left\{ {\sum_{(s_{1},\dots,s_{m+k+1})}} \frac{\Gamma(\xi + s_{m+k+1})}{s_{m+k+1}!}\prod_{j=1}^{m+k+1} \frac{\Gamma\left(j+\sum_{i=1}^{j}s_{i}\right)}{\Gamma\left(j+\sum_{i=1}^{j-1}s_{i}\right)}\right\}. $$ $(s_{1},\dots ,s_{m+k+1})$ under the sum sign indicates summation over all (ordered) partitions of $s$ into $m+k+1$ non-negative integers, $s_{1},\dots ,s_{m+k+1}$. Order matters here, i.e., $(1,0)$ is different from $(0,1)$.
Numerical calculations of the coefficients (with highest possible precision on my laptop) show excellent match (5 or more digits) with the formula, even for exotic indices, like $\mu=\pi$ and orders up to $y^{-6}$.
For the OP's functions I get: $$ C(x) e^{-x^{2}} = 1 + \frac{3}{8} x^{-2} + \frac{65}{128} x^{-4} + \frac{1225}{1024} x^{-6} + \frac{131691}{32768} x^{-8} + O(x^{-10}) , $$ $$ L(x) x^{2} e^{-x^{2}} = 1 + \frac{15}{8} x^{-2} + \frac{665}{128} x^{-4} + \frac{19845}{1024} x^{-6} + \frac{2989371}{32768} x^{-8} + O(x^{-10}) . $$
All calculations were done with Mathematica 11.