Can a non-regular language be made regular via concatenation when they don't share characters?
First, the equality $L = AB$ is correct. Next, if you know that $B$ is non-regular, you can prove that $L$ is non-regular as follows. 

Suppose that $L$ is regular. Then $(00)^{-1}L = \{0^n1^m2^m \mid n \geqslant 0, m&gt;1\} = 0^*B$ is regular. It follows that $0^*B \cap \{1,2\}^* = B$ is regular. Contradiction.