Any regular language upholds the pumping lemma. Hence, if a language doesn't uphold the pumping lemma, it is not a regular language. This is in fact the standard way of proving a language is not regular.
Let us choose $p$ to be the pumping length. Let's now take $0^n=1^n \oplus 1^n$, which is in $A$. This string can be decomposed into three concatenated substrings $xyz$. Since $|xy| \leq p$, we see that $v=0^k$ for some $k$ greater than 0. If we choose to pump it $i$ times, we will then obviously obtain an $x$ which is longer ($i > 1$) or shorter ($i = 0$) than the other two operands and cannot then be the result of their XOR, hence proving that the language is not regular.
However, it may be that you forgot to add some clauses to the problem statement, since this seems quite trivial. Edit: my intention was not to point out that this is easy in general, the principle itself of the proof being rather non-trivial. However, some additional clauses, such as the strings having the same length would render this sensibly harder, and it may be what the problem had in mind.