k-query oracle Turing machine
Let's try to show the contrapositive. Since $\mathsf{NP} \cup \mathsf{coNP} \subseteq \mathsf{P}^{\mathsf{SAT},1}$, the contrapositive states that if $\mathsf{P}^{\mathsf{SAT},1} \subseteq \mathsf{NP} \cup \mathsf{coNP}$ then $\mathsf{NP} = \mathsf{coNP}$. Let us assume, therefore, that every problem in $\mathsf{P}^{\mathsf{SAT},1}$ is both in $\mathsf{NP}$ and in $\mathsf{coNP}$. We would like to show that this implies that $\mathsf{NP} = \mathsf{coNP}$.
We can imagine that the argument will have the following structure:

Come up with some language $L \in \mathsf{P}^{\mathsf{SAT},1}$.
The assumption implies that $L \in \mathsf{NP}$ or $L \in \mathsf{coNP}$.
Use these facts to say something about the relation between $\mathsf{NP}$ and $\mathsf{coNP}$.

Our goal is to show that $\mathsf{NP} = \mathsf{coNP}$, or equivalently $\mathsf{NP} \subseteq \mathsf{coNP}$. This, in turn, is equivalent to showing that $\mathsf{SAT} \in \mathsf{coNP}$. Considering the plan above, what we need is to find a language $L \in \mathsf{P}^{\mathsf{SAT},1}$ such that both $L \in \mathsf{NP}$ and $L \in \mathsf{coNP}$ imply $\mathsf{SAT} \in \mathsf{coNP}$.
We can take as $L$ the following language:
$$
L = \{(x,0) : x \in \mathsf{SAT}\} \cup \{(x,1) : x \notin \mathsf{SAT}\}.
$$
Details left to you.