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.
No, this is not possible in general. $A^{LS}$ might have trivial intersection with a non-zero hereditary $C^\ast$-subalgebra of $A$, and thus any non-zero positive element in such a hereditary $C^\ast$-subalgebra cannot be approximated from below by elements in $A^{LS}$.
For a concrete example, I will give a non-unital example. You can simply unitise this example to get a unital counter-example (but I leave this to the reader).
Consider $\mathbb C^n \subseteq \ell^2(\mathbb N)$ in the obvious way, and let $A = \mathcal K(\ell^2(\mathbb N))$ with finite dimensional $C^\ast$-subalgebras $A_n = \mathcal K(\mathbb C^n)$. Fix any unit vector $\xi \in \ell^2(\mathbb N) \setminus \bigcup_n \mathbb C^n$ and let $p$ be the projection onto the span of $\xi$. Then $pAp \cap \bigcup_n A_n = \{0\}$, and thus $0$ is the only positive element in $\bigcup A_n$ which is below $p$.
Best Answer
There are no solutions, even if we only assume that $f$ has a derivative at every point.
Note that if $f(x)>x$ for all $x$, then $f(f(x))>f(x)>x$ etc, thus $\sum a_k f^k(x)\geqslant (\sum a_k)x$ that is positive for $x>0$. Analogously, if $f(x)<x$ for all $x$, then $f(f(x))<f(x)<x$ etc, and $\sum a_k f^k(x)\leqslant (\sum a_k)x$ that is negative for $x<0$. Thus, by Intermediate Value Theorem, there exists $x_0$ such that $f(x_0)=x_0$. If $p=f'(x_0)$, taking the derivative of our equation at $x_0$ we get $\sum a_kp^k=0$, thus, $f(x)=px$ is a solution of Eq.