Complexity of advice language?
Rigorous and probably more correct version at the bottom.

I don't think so. Consider the case where A(L,M) consists of advice strings of even length. To decide if x is in L, the machine M gets advice yz where y and z are strings of equal length. M discards z and decides membership of x in L only depending on y. Let y be of length n, and let z range over some arbitrary subset S of strings of length n.

In such a case, A(L,M) cannot be in P/Poly since given a string yz, the machine has to test membership of z in S. But since the advice is of length poly(n), the machine cannot work correctly for every S by counting argument.

Another more rigorous version:

Let L be the trivial language containing all strings. The machine M ignores its advice and accepts every input. The task is now to construct the set of advice strings A(M) which cannot be in P/Poly subject to the constraint that strings in L of length $n$ get advice of length $O(poly(n))$.

For every length $i$, let $S_i \subseteq {0,1}^i$. Let $S = \cup_i S_i$. Construct a mapping $f: S \rightarrow Z^+$ such that $|s| \leq f(s)$. The mapping just enumerates the strings in $S$ in non-decreasing order of length. The string $s$ serves as advice to M for strings of length $f(s)$. 

By choosing $S_i$ to be of superpolynomial description size (can be done because of counting arguments), we get that $S$ (which is also $A(M)$) lies outside P/Poly.