Has it been shown or can we show that if $SAT \in P$ then SAT can't be in any complexity class C so that $C \subsetneq P$?
SAT is NP-hard with respect to NC$^0$ reductions, and in particular, it is P-hard with respect to NC$^0$ reductions.
Now let $C$ be a complexity class which is closed under NC$^0$ reductions. If SAT is in $C$ then every problem in P can be reduced to $C$ using NC$^0$ reductions, and so belongs to $C$, implying that $C$ contains P.