Graph isomorphisim is not proven to be NP-complete what would it imply if it were possible to prove that there are some problems which are in NP set of problems but not in NP-complete set.
As written, the question is a bit trivial: if NP = NP-complete, then since P $\subseteq$ NP we get P=NP since every problem in P would be NP-complete.
I suspect what's meant, though, is the following:
Suppose there are no NP-intermediate problems; that is, that every problem in NP is either in P or is NP-complete. What does that tell us about P vs. NP?
This is definitely not trivial, but the answer turns out to be the same: Ladner's theorem says that if there are no NP-intermediate problems then P=NP.
It's worth noting that Ladner's theorem is constructive in that Ladner provides a specific example of a language $X$ such that if P$\not=$NP then $X$ is NP-intermediate. (I'm not making the stronger claim that it doesn't need the excluded middle, although I suspect that's true.) Moreover, the proof is pretty short; see e.g. here.