Does NP=coNP imply P=NP?
The case that $P \subset NP = coNP$ is possible. The reason why $P = NP$ implies $NP = coNP$ is that $P$ is closed under complement, and so if $P = NP$, $NP$ must be closed under complement as well.

A class being closed under complement doesn't necessarily imply anything about its deterministic counterpart. For example, we know that $NL$ is closed under complement, but don't know whether $L = NL$.