Is $P^{NPI}$ different from $P^{NP}$?
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).