This is example I-22 in Eisenbud's Geometry of Schemes.
Let $K$ be a field and $R=K[x]_{(x)}$ the localization of $K[x]$ at the maximal ideal (x). So $R$ is basically $K(x)$ except the elements where the denominator has a non-zero constant term. In the book it is written that $Spec(R)=\{(0), (x)\}$, but how can $(x)$ be a prime ideal of $R$, since it clearly contains $1$ and hence is equal to $R$, which is by definition not a prime ideal?
The ideal $xR$ is proper, and you can prove it directly: If $xf(x)/g(x) = 1$ with $f(x), g(x) \in K[x]$, then $xf(x) = g(x)$; however, as you said, in $R$ our denominators must have non-zero constant terms.
For algebraic geometry, it's important to know that $\mathfrak{p} \mapsto S^{-1}\mathfrak{p} = \mathfrak{p}(S^{-1}A)$ is a bijection between the primes of $A$ not meeting $S$ and the primes of $S^{-1}A$. This is Proposition 6.4 of Milne's notes and should be done in any commutative algebra textbook. [I think Atiyah-Macdonald is still horribly expensive, but that's the one I'd recommend.]