I checked if anything on this questions exists on MSE, but I found nothing. I did find two solutions available via google, but was skeptical of both.
This is Exercise 21 - i) from Chapter 3 of Introduction to Commutative Algebra by M.F. Atiyah and I.G. Macdonald.
The exercise states:
Let $A$ be a ring, $S$ be a multiplicatively closed subset of $A$, and $\phi \colon A \to S^{-1}A$ the canonical homomorphism. Show that $\phi^{*} \colon \mathrm{Spec}(S^{-1}A) \to \mathrm{Spec}(A)$ is a homeomorphism of $\mathrm{Spec}(S^{-1}A)$ onto its image in $X = \mathrm{Spec}(A)$. Let this image be denoted by $S^{-1}X$.
I am satisfied with 3/4'ths of my solution. Namely, we know that $\phi^{*}$ is continuous by Exercise 21 - i, Chapter 1. We also know this restriction mapping is surjective, trivially. Further $\phi^{*}$ is injective by applying Exercise 20 - ii, Chapter 3 that every prime ideal of $S^{-1}A$ being an extended ideal is a sufficient condition for infectivity. We know the hypothesis that every prime ideal of $S^{-1}A$ is extended is satisfied since Proposition 3.11 - iv) gives a bijective correspondence, $\mathfrak{p} \longleftrightarrow S^{-1}\mathfrak{p},$ where $\mathfrak{p}$ are the prime ideals of $A$ that don't meet $S$ and $S^{-1}\mathfrak{p}$ represents $\mathfrak{p}^{e}$, thus every prime ideal in the localization is extended.
My issue is with the final step, showing that the inverse map is continuous. To be honest, I have been struggling to decided what precisely the rule defining the inverse function would be? I believe it is simply the action of $\phi$? I decided to follow the same outline for how this text had us show that $\phi^{*}$ it self was continuous, so here is the proof I came up with. I am very skeptical however as I don't believe I am using the follow power of my hypothesis:
Proof: To show ${\phi^{*}}^{-1}$ is continuous we show ${\phi^{*-1}}^{-1}(X_{f}) = Y_{\phi^{-1}(f)}$, apologies for the atrocious notation on the left hand side, I simply mean the pullback of the open basis set $X_f$ in $\mathrm{Spec}(S^{-1}A)$ under the inverse of $\phi^{*}$. If successful this obviously satisfies the usual topological definition of continuity since $Y_{\phi^{-1}(f)}$ is open in $S^{-1}X \subseteq \mathrm{Spec}(A).$ The argument is
$$\mathfrak{p} \in {\phi^{*-1}}^{-1}(X_{f}) \iff \phi^{*-1}(\mathfrak{p})\in X_{f}$$ $$\iff \phi(\mathfrak{p}) \in X_{f}$$
but since $\mathfrak{p}$ is some contracted ideal, $\phi(\mathfrak{p})$ is just some $\mathfrak{q}$ in $S^{-1}X$, then
$$\iff f \notin \phi(\mathfrak{p})$$
$$\iff \phi^{-1}(f) \notin \mathfrak{p}$$
$$\iff \mathfrak{p} \in Y_{\phi^{-1}(f)}.$$
Let $Y_g$ be some distinguished open set of $Spec(S^{-1}A)$. Then $g\in S^{-1}A$ so we can write $g=h/s$ for $h\in A$ and $s\in S$. But $s$ is a unit in $S^{-1}A$ so actually $Y_g=Y_{h/s}=Y_{h/1}$. Then, you can check that the image of $Y_{h/1}$ under $\phi^*$ is given by $X\cap X_h$, which is open in $X$. This shows that $\phi^*$ is an open map.