For a localization of a ring $S^{-1}R,$ the extension of a prime ideal $P^e$ has no units if $P \cap S = \emptyset.$

Question

In my lecture notes, my professor wrote down that if a prime ideal $P$ of $R$ is such that $P \cap S = \emptyset$ then the extension of $P$ via a fixed homomorphism $\phi: R \rightarrow S^{-1}R$ contains no units as well. This confuses me. I know that $\phi(S) \subset (S^{-1}R)^\text{x}$ but couldn't $P^e = \phi(P)(S^{-1}R)$ contain a unit not in $\phi(S)$ then?