Is there a well-known nontrivial subset of NP for which it is known that it is a proper subset?
[Answering the original motivating issue: "Should I just accept the first statement since NPC is defined with respect to deterministic polytime reductions?"]

First, if $P=NP$ but the Berman-Hartmanis Isomorphism Conjecture fails, then there are infinitely many distinct p-isomorphism classes of languages within $P$. So even though it is true that $P=NP$ implies all sets in $P$ are $NP$-complete with respect to polynomial-time reduction (technical caveat below), that does not mean that we lose all information about $P$. By looking at a finer type of reduction -- but still polynomial-time, namely polynomial-time computable-and-invertible length-increasing one-one reductions -- we can potentially tell a lot about what's going on inside $P$.

In fact, if $P=NP$ but the Isomorphism Conjecture fails, then the partially ordered set of p-isomorphism classes inside $P$ is quite rich: any countable partial order can be embedded in it (see P. Young. Some structural properties of polynomial reducibilities and sets in NP. STOC 1983.). 

But more generally, in some cases it does feel wrong to lump all of $P$ together and in other cases it's exactly what you want. This is why, for example, we don't use poly-time reductions when discussing $L$ versus $P$, but we do when discussing $P$ versus $NP$. But there might be another class you want to contrast with $NP$ where it makes sense to use other reductions. The type of reduction you use should be the type you need for the problem at hand; there is no "single right reduction" to define, e.g., $NP$-completeness (though "$NP$-complete" without qualification usually means w.r.t. poly-time many-one reductions, but that is a matter of convention).

Distinctness of complexity classes is not parametrized by reductions; however, there are several potentially distinct classes that could legitimately be called "$NP$-complete," namely for any reducibility $r$, "$NP$-complete with respect to $r$-reductions".

(Caveat: If we use poly-time many-one reductions, then $\emptyset$ and $\Sigma^{*}$ do not become $NP$-complete if $P=NP$. If we use poly-time Turing reductions, or even poly-time 1-tt reductions, they do.)