There is no mistake; the one-to-one, inclusion preserving correspondence between ideals of a ring $R$ that contain the ideal $I$ and ideals of the quotient ring $R/I$ also gives a correspondence between prime ideals.
For the commutative-with-unity case, your proof works.
If you want a proof for the more general case, and to prove a bit more:
Definition. Let $R$ be a ring; an ideal $I$ of $R$ is completely prime if and only if whenever $ab\in I$, either $a\in I$ or $b\in I$.
Definition. Let $R$ be a ring; an ideal $I$ of $R$ is prime if and only if whenever $JK\subseteq I$, where $J$ and $K$ are ideals, either $J\subseteq I$ or $K\subseteq I$.
If $R$ is commutative, then an ideal is prime if and only if it is completely prime. The two notions are not equivalent for arbitrary rings, though. For example, in the ring of $2\times 2$ matrices over a field, the only ideals are the trivial ideal and the whole ring; in particular, the zero ideal is a prime ideal, but it is not completely prime because you can find two nonzero matrices whose product is the zero matrix.
Theorem. Let $R$ be a ring and let $I$ be an ideal of $R$. Then the natural correspondence between ideals of $R$ that contain $I$ and ideals of $R/I$ identifies prime ideals with prime ideals and identifies completely prime ideals with completely prime ideals.
Proof. Let $J$ be an ideal that contains $I$. If $J$ is completely prime, suppose that $a+I, b+I\in R/I$ are such that $(a+I)(b+I)\in J/I$. Then $ab+I\in J/I$, hence $ab\in J$ (since $I\subseteq J$); since $J$ is completely prime, either $a\in J$ or $b\in J$, so either $a+I\in J/I$ or $b+I\in J/I$. Thus, $J/I$ is completely prime. Conversely, suppose that $J/I$ is completely prime, and let $a,b\in R$ be such that $ab\in J$. Then $ab+I\in J/I$, hence either $a+I\in J/I$ or $b+I\in J/I$. If $a+I\in J/I$, then there exists $i\in I$ such that $a+i\in J$, hence $a\in J$ (since $I\subseteq J$); likewise, if $b+I\in J/I$, then $b\in J$. Thus, $J$ is completely prime.
Now suppose that $J$ is a prime ideal. Let $K/I$ and $L/I$ be ideals of $R/I$ such that $(K/I)(L/I)\subseteq J/I$, with $K/I$ corresponding to the ideal $K$ of $R$ that contains $I$, and the ideal $L/I$ corresponding to $L$. Then $(K/I)(L/I) = (KL)/I\subseteq J/I$, hence, by the inclusion-preserving correspondence, $KL\subseteq J$, so either $K\subseteq J$ or $L\subseteq J$, hence $K/I\subseteq J/I$ or $L/I\subseteq J/I$. Thus, $J/I$ is a prime ideal. Conversely, if $J/I$ is a prime ideal, let $K$ and $L$ be ideals such that $KL\subseteq L$. Then $K+I$ and $L+I$ are ideals that contain $I$, and $(K+I)(L+I)=KL+KI+IL+I^2\subseteq KL+I\subseteq KL+J=J$; therefore, $(K+I)/I$ and $(L+I)/I$ are ideals of $R/I$ whose product is contained in $J/I$, so either $K+I\subseteq J$ or $L+I\subseteq J$. But $K\subseteq K+I$ and $L\subseteq L+I$, so either $K\subseteq J$ or $L\subseteq J$. Thus, $J$ is prime. QED
The homomorphisms $A \to A/I$ and $A \to S^{-1}A$ induce continuous maps of spectra in the other direction. These are injective and you know the images. You want them to be homeomorphisms onto their images.
There's probably a better way to think about this, but given an ideal of $A/I$ defining a closed set there, what do you think the associated ideal of $A$ should be?
For an example, remember that the most common multiplicative sets are $S = \{1, f, f^2, \dots\}$ and $S = A - \mathfrak{p}$. The first kind always produces an open subset, but the second need not. Try looking in $\operatorname{Spec} \mathbb{Z}$.
Best Answer
Consider the ideals $(x)$ and $(xy)$ in the ring $k[x,y]$ and its localization $k[x,y]_{(x)}$.