Oracle Turing Machine EXP^EXP
No, $\mathsf{EXP^{EXP}=2EXP}$, a set of languages decidable in $O\left(2^{2^{\mathrm{poly}(n)}}\right)$ time.

This is just because you can give exponentially long input to an oracle which can solve it. So, the total power is $\exp(\exp(n))\ne \exp(n)$.

To see why $\mathsf P$ is self-low just take a machine that can run quadratic time and give to it the same oracle. While the resulting machine will be able to solve all languages in $\mathsf{DTIME}(n^4)$ it is still polynomial. Generalizing, $\mathrm{poly}(\mathrm{poly}(n))=\mathrm{poly}(n)$.

Was wrong, now understood that this is the only way to make machine that faster. Thanks to Ariel.

P.S. There is a parsimonious exp-time reduction from $\mathsf{2EXP}$-complete problem to an exponentially longer $\mathsf{EXP}$-complete problem.