$\mathsf{NP} \cap \mathsf{coNP}$ as oracle Does $\mathsf{NP^{NP \,\cap\, coNP}=NP}$ hold?

Clearly $\mathsf{NP^{NP}\neq NP}$, but it seems to me that $\mathsf{NP\cap coNP}$ is "deterministic" which makes me believe this is true. 

Is there a simple proof (or maybe just by definition)?
Yes. Indeed, an oracle $A$ satisfies $\mathsf{NP}^A=\mathsf{NP}$ if and only if $A \in \mathsf{NP} \cap \mathsf{coNP}$. This class is called $\mathsf{Low(NP)}$ or sometimes $\mathsf{L_1 P}$ (see the link and the paper cited there for more of an explanation of the low hierarchy in general).

Your intuition about "determinism" is actually somewhat correct (although it's not deterministic enough for us to conclude $\mathsf{P} = \mathsf{NP} \cap \mathsf{coNP}$). Try this as an exercise and you'll see this intuition vindicated: first show - carefully, spelling out the details - that if $A \in \mathsf{P}$, then $\mathsf{NP}^{A} = \mathsf{NP}$. Figure out exactly the part of your proof that doesn't work if you only assume $A \in \mathsf{NP}$, and then realize why that part does work when $A \in \mathsf{NP} \cap \mathsf{coNP}$.

(Showing the converse is not too difficult either: $\mathsf{NP}^A = \mathsf{NP}$ implies $A \in \mathsf{NP} \cap \mathsf{coNP}$.)