If $x \in L$ only if $x \in A$ and $x \in B$, where A is an NP problem and B is a coNP problem, I cannot place $L \in NP$ or $L \in coNP$ without implying that NP = coNP right?
No, because $A$ isn't necessarily $NP$-complete and $B$ isn't necessarily $coNP$-complete. You'd need both to be true to force the implication $NP = coNP$ by $X$ being a member of both problem classes. In fact $A$ and $B$ could both be in $NP$ and $coNP$. E.g. $A$ and $B$ could both be in $P$. $NP \cap coNP$ is not the empty set.