In Dummit & Foote, Proposition $46$(2) on page $718$ says
If $R$ is an integral domain, then $R_P$ is an integral domain. The ring $R$ injects into the local ring $R_P$, and, identifying $R$ with its image in $R_P$, the unique maximal ideal of $R_P$ is $PR_P$.
Let $\varphi: R \to R_P$ be the canonical map, where $R$ is not an integral domain. Is it not true in general that $P^e (= \varphi(P)R_P)$ is not equal to $PR_P$?
If $\varphi: R \to S$ is a ring homomorphism of commutative rings, then the extension of an ideal $I \subset R$ is the ideal $\varphi(I)S$, and the contraction of an ideal $J \subset S$ is the ideal $\varphi^{-1}(J)$.
In the special case where $R$ is a subring of $S$ and $\varphi$ is the natural injection, the extension of $I \subset R$ is written as $IS$, and the contraction of $J \subset S$ is written as $J \cap R$.
If $R$ is an integral domain and $P$ is prime, then the natural map $R \to R_P$ is injective and we can identify $R$ as a subring of $R_P$. Then the maximal ideal of $R_P$ is $PR_P$ by the special case above.
In general, since $P^e$ is the ideal of $R_P$ generated by all products $\frac{a}{1}\frac{r}{u}$ where $a \in P$, $r \in R$, $u \notin P$, then this is the same as the ideal $PR_P$. But it seems Dummit & Foote reserve the notation $PR_P$ for when $R$ injects into $R_P$.