It is more common to just write $L_g$ for $L_{x_g}$.  As $L(G)$ admits a finite trace, there is a natural injective map $L(G)$ into $\ell^2(G)$-- this is your map $A \mapsto (\mu_g)$.  It is absolutely not true that this map surjects (Open Mapping Theorem).  It is obviously sufficient that $(\mu_g)\in\ell^1(G)$ for there to be some $A$ giving rise to $(\mu_g)$.

With $G=\mathbb F_2$, one can say a bit more.  For example, Haagerup showed in:<br/>
Haagerup, Uffe<br/>
An example of a nonnuclear C∗-algebra, which has the metric approximation property.<br/>
Invent. Math. 50 (1978/79), no. 3, 279–293. <br/>
See Lemma 1.4 that if $f$ is a function of finite support, then denoting $f_n$ the function which agrees with $f$ on the collection of words of reduced length $n$, and is zero elsewhere, we have that there is $A\in L(G)$ inducing $f$, with $\|A\| \leq \sum_{n\geq 0} (n+1) \|f_n\|_2$.  From this, it's easy to construct functions not in $\ell^1(G)$, but which are nonetheless induced by members of $L(G)$.