What does it mean when $A$ is a NP-Complete Problem but $\bar{A} = NP$?
If $A$ is ${\sf NP}$-complete and $\bar A$ is in $\sf NP$ then we can show that ${\sf NP = coNP}$. This would imply that the polynomial hierarchy collapses to the first level, which implies ${\sf NP = coNP = PH}$. 

However, this implies not that all problems are in NP. Just an easy example. There are problems like the halting that are undecidable, and they will remain undecidable even if ${\sf NP = coNP}$, since the proof that the halting problem is undecidable does not use the concept of $\sf NP$ at all.