Is $P^{NPI}$ different from $P^{NP}$? Can we prove that for every language $L\in\mathsf{NP}$ that is not $\mathsf{NP}$-hard (this assumes $\mathsf P \ne \mathsf{NP}$), $\mathsf{P}^L \ne \mathsf{P}^{\text{SAT}}$?  Alternately, can this be proven under any reasonable assumptions?
Depends on your definition of NPI. If A is incomplete for Turing reductions, the answer is yes since SAT is not in $P^A$.

If A is just many-one incomplete then we don't know how to prove it. We have a relativized world with there is a set A in NP such that A is not NP-complete via many-one reductions but SAT can be computed by a single query to A. (Theorem 1.9 in this paper).