 Solution

The image $f/1$ is a unit in the localization at a prime ideal $R_{\mathfrak{p}}$, because $f\notin \mathfrak{p}$. The universal property of localization, applied to the ring homomorphism $R\to R_{\mathfrak{p}}$ and the multiplicative subset $\{1,f,f^2,\dots \}$, gives a unique ring homomorphism fixing the image of $R$:Fixing the image of $R$ is exactly the required compatibility with the algebra over a ring structures. Uniqueness also follows directly: such a ring homomorphism must send $1/f$ to the inverse of $f/1$, and these elements together with $R$ generate $R_f$.