Proving that if A⊕B ∈ NP then NP = coNP
Thanks to G. Bach and Yuval Filmus.

If every $A, B \in NP$ and $A \oplus B \in NP$, then let $E$ to be some language $E\in NP$. Since $\Sigma^*\in NP$ then $E \oplus \Sigma^* \in NP$. But $ E \oplus \Sigma^* = \bar E$, so $\bar E \in NP$, and we got that $NP \subseteq coNP$.

Let $D$ to be some language $D\in coNP$ ($\bar D \in NP$). Since $\Sigma^*\in NP$ then $\bar D \oplus \Sigma^* \in NP$. But $ \bar D \oplus \Sigma^* = D$, so $\bar D \in NP$, and we got that $coNP \subseteq NP$.

And then $NP = coNP$.