P vs NP and can an oracle make P=EXPTIME?
First let me try to be a bit more precise about your first sentence.  The reasoning you list shows that no relativizing proof can prove or disprove P=NP.  "Diagonalization" is a fuzzy concept, and depending on what you mean by it, might or might not relativize -- diagonalization proofs tend to relativize, but no guarantees.  So the oracle result rules out some/many kinds of diagonalization proofs, but maybe not all.  See, e.g., https://cstheory.stackexchange.com/q/6575/5038.

Second, let's discuss your final question.  Yes, it is indeed true that $P^X \ne EXPTIME^X$ for all $X$; there is no oracle $X$ that will make them equal.  I'll talk through why that is.  The standard way to prove $P \ne EXPTIME$ is to prove $TIME(p(n)) \ne TIME(p(n)^2)$ (the time hierarchy theorem), then notice that this implies $P \ne EXPTIME$ (since $TIME(2^n) \ne TIME(2^{2n})$, but $P \subseteq TIME(2^n)$ and $TIME(2^{2n}) \subseteq EXPTIME$).  The proof of $TIME(p(n)) \ne TIME(p(n)^2)$ uses diagonalization.  This proof does indeed relativize.  In particular, for all oracles $X$, we do indeed have $TIME^X(p(n)) \ne TIME^X(p(n)^2)$.  It follows that, for all oracles $X$, we have $P^X \ne EXPTIME^X$ (by the same argument as before).

See also: https://cstheory.stackexchange.com/q/8902/5038, https://cstheory.stackexchange.com/q/6575/5038.