Could the empty language be NP-Complete?
You are correct that $\emptyset\in NP$, since we already know that $\emptyset$ can be decided in constant time (a TM that immediately rejects), and $DTIME(O(1))\subseteq P\subseteq NP$.
But $\emptyset$ is not NP-complete, regardless of whether $P=NP$. Indeed, by definition, a NP-complete language $A$ is such that every language $B \in NP$ admits a Karp reduction to $A$, i.e., a function $f$ that is computable in polynomial time and such that $x \in B \iff f(x) \in A$.
If $A= \emptyset$ and $B \neq \emptyset$ then no such function exists, since there is no way to satisfy $x \in B \implies f(x) \in A$.