I think that while the language may not be context-free (and it certainly doesn't look context-free), the pumping lemma cannot be used to prove it. It seems to me that you have put your finger on the difficulty. Let $s\in A$. Then $s=a^nb^na^m$ with $n\neq m.$ In the case that $n>m$ we can write $s=uvxyz$ where $$\begin{align} u&=a^n,\\v&=a,\\x&=\varepsilon,\\y&=b,\\z&=b^na^m\end{align}$$ Then for $i\geq0$,$$uv^ixy^iz=a^na^ib^ib^na^m=a^{n+i}b^{n+i}a^m\in L,$$ since $n+i>m$.
On the other hand, if $n<m$, we can write $s=uvxyz$ where $$\begin{align}
u&=a^nb^na^m,\\
v&=a,\\
x&=\varepsilon,\\
y&=\varepsilon,\\
z&=\varepsilon
\end{align}$$ and now for $i\geq0$,$$uv^ixy^iz=a^nb^na^ma^i=a^nb^na^ma^i=a^nb^na^{m+i}\in L,$$ since $n<m+i$.
Thus there is no string that we use can use for a pumping lemma proof.
The Wikipedia article on the pumping lemma states that there are alternatives to the pumping lemma for proving a language is not context-free. They mention Ogden's lemma and the interchange lemma, but I never studied either of these, so I can't be of much help. (Or not until I read up on them, at least.)
I found this example of the use of Ogden's lemma, which may be of use to you. I haven't found out enough about Ogden's lemma yet to understand the example myself.
