Show that $R\cong R_P$, the ring of quotients of $R$ with respect to the multiplicative set $R-P$ if $R$ has exactly one prime ideal $P$.

Question: Let $R$ be a commutative ring with identity that has exact one prime ideal $P$. Show that $R\cong R_P$, the ring of quotients of $R$ with respect to the multiplicative set $R-P$

This is an old question, and the first part of the question was to show that $R/P$ is a field. So, we consider maximal ideal $M$ of $R$ such that $P\subseteq M\subsetneq R$. Thus, $M$ is prime and so $M=P$ by uniqueness of $P$. Thus $P$ is maximal and so $R/P$ is a field. Now, for the question, I'm not sure if it is relevant, but whenever I see "multiplicative set", I think of the theorem:

Let $R$ be a nonzero commutative ring with identity and $S$ a multiplicative subset of $R$ not containing $0$. If $P$ is maximal in the set of ideals of $R$ not intersecting $S$, then $P$ is prime.

But, I don't know if this will help. To show the isomorphism, I imagine I need to come up with a map, but I can't quite see one that would work. Any help is greatly appreciated! Thank you.

Since maximal ideals are prime, there is also at most one maximal ideal. And since there exists a maximal ideal (unless $R=0$, in which case there would be no prime ideal), there is exactly one maximal ideal, which is at the same time the unique prime ideal, making $R$ a local ring.

And it is known that the localisation of a local ring at its maximal ideal is just the local ring.