Is P^SAT with only one query equal to the union of NP and coNP?
Under the assumption that $NP\neq \mathrm{co}NP$, we have that:

$$P^{SAT[1]}=NP\cup \mathrm{co}NP\implies P^{NP[1]}\neq P^{NP[2]}$$

Proof: Since $D^p\subseteq P^{NP[2]}$, we deduce that $(SAT\dot{\land} UNSAT)\in P^{NP[2]}$.

Meanwhile, $(SAT\dot{\land}UNSAT)\notin NP$ due to $UNSAT\notin NP$ and similarly, $(SAT\dot{\land}UNSAT)\notin \mathrm{co}NP$ due to $SAT\notin \mathrm{co}NP$. So, $(SAT\dot{\land}UNSAT)\notin NP\cup \mathrm{co}NP=P^{NP[1]}$.

We conclude that $P^{NP[1]}\neq P^{NP[2]}$.$\blacksquare$

In the above, $(SAT\dot{\land}UNSAT)=\{(\phi_1,\phi_2)\mid\phi_1\in SAT \land \phi_2\in UNSAT\}$.

So, though it might be tempted to push $P^{NP[1]}$ downto $NP\cup \mathrm{co}NP$, this result (if true) could tell us more about where the polynomial-time hierarchy $PH$ would actually collapse to. And, in this case, your proposition would save $PH$ from collapsing to $P^{NP[1]}$.