The form of prime ideals in $S^{-1}R$ 
Theorem. Let $R$ be a commutative ring with $1_R$, $P\in \mathrm{Spec}(R)$ a prime ideal of $R$,  $S\subseteq R$ an
  multiplicative subset of $R$ and $\nu:R\longrightarrow S^{-1}R,\ a
 \longmapsto \nu(a):=\frac{a}{1_R}$ the natural homomorphism. Then:
I. a) If $P\in \mathrm{Spec}(R)$ and $P\cap S =\emptyset$, then
  $P^e\in \mathrm{Spec}(S^{-1}R)$.
b ) If $P'\in \mathrm{Spec}(S^{-1}R)$, then $(P')^c\in
 \mathrm{Spec}(R)$ and $(P')^c\cap S=\emptyset$.
ΙΙ. Every prime ideal of $S^{-1}R$ is in the form $P^e$, for some
  unique prime ideal $P\in \mathrm{Spec}(R)$ of $R$ which has the
  property $P\cap S=\emptyset$. That is, 
  $$\boxed{\mathrm{Spec}(S^{-1}R)=\{ I\in \mathcal{ID}(S^{-1}R):
 \exists! P\in \mathrm{Spec}(R) \text{ s.t. } I=P^e \text{ and } P\cap
 S=\emptyset\} } .$$

Proof. I. a) Let $P\in \mathrm{Spec}(R)$ and $P\cap S =\emptyset$. We will show that $P^e\in \mathrm{Spec}(S^{-1}R)$.
Indeed:


*

*We have $P^e \neq S^{-1}R$, because if we suppose contrary that $P^e\ = S^{-1}R$, then $P^e\ = S^{-1}R\implies (P^e)^c\ = (S^{-1}R)^c \iff P^{ec}\ = (S^{-1}R)^c \implies P = (S^{-1}R)^c$ (and the last implication becomes from a proposition in which under this hypothesis $P^{ec}=P$).
But, we observe that
\begin{alignat*}{2}
(S^{-1}R)^c \quad = \quad &\nu^{-1}(S^{-1}R) \\
 \quad = \quad & \{ x\in R: \nu(x)\in S^{-1}R  \} \\
 \quad = \quad & \Big\{ x\in R: \frac{x}{1_R} \in S^{-1}R  \Big\} \\
 \quad = \quad & R.
\end{alignat*}
Thus,  $P=(S^{-1}R)\iff P=R$, conctradiction, because from hypothesis  $P \in \mathrm{Spec}(R)$.

*Let's take two elements $\alpha:=\frac{r}{s}\in S^{-1}R$ and $\alpha':=\frac{r'}{s'}\in S^{-1}R$ s.t. $\alpha \alpha'\in P^e$. Then, from the well known proposition we take 
\begin{alignat*}{2}\alpha \alpha'=\frac{r}{s}\cdot \frac{r'}{s'}=\frac{rr'}{ss'}\in P^e  \implies & rr'  \in P 
  \implies  r\in P  \text{ or } r'\in P \\
\implies & \frac{r}{s}\in P^e \text{ or } \frac{r'}{s'}\in P^e \\
\iff & \alpha \in P^e \text{ or } \alpha'\in P^e,
\end{alignat*}
so we get $P^e\in \mathrm{Spec}(P^{-1}R)$, as we wanted.
b) Let $P'\in \mathrm{Spec}(S^{-1}R)$ a prime ideal of $S^{-1}R$. 
1) We will show that $(P')^c \in \mathrm{Spec}(R)$. One can easily prove that a contraction of a prime ideal is again a prime ideal. So, $P'\in \mathrm{Spec}(S^{-1}R) \implies (P')^c \in \mathrm{Spec}(R)$. $\checkmark$
2) Now, we will show that $(P')^c\cap S=\emptyset$.
First of all, it is a well known lemma which states that $\mathcal{ID}(S^{-1}R)=\mathcal{E}(S^{-1}R) $. So, for the ideal$P'\trianglelefteq S^{-1}R$ we conclude that
$$P' \in \mathcal{ID}(S^{-1}R)=\mathcal{E}(S^{-1}R) \iff \exists I \trianglelefteq R:\ P'=I^e.$$
Then, it is
$$ (P')^c=I^{ec} \iff (P')^{ce}=I^{ece} \iff (P')^{ce}=I^{e} \iff (P')^{ce}=P'.$$
We contrary suppose that $(P')^c\cap S \neq \emptyset$.
Then, there is some $x \in (P')^c\cap S \iff x\in (P')^c$ and $x\in S$. 
We observe that the element $x\in (P')^c\trianglelefteq R$ and simultaneously $x\in S$. Sot the element $\frac{x}{x}\in ((P')^c)^e=(P')^{ce}$ and from this we take
$$1_{S^{-1}R}=\frac{1_R}{1_R}=\frac{x}{x} \in (P')^{ce} = P' \iff P'=S^{-1}R.$$
Contradiction! Because we supposed that $P'\in \mathrm{Spec}(S^{-1}R)$.
Finally, $(P')^c\cap S = \emptyset$. $\checkmark$
Questions.
1) Could you please verify me if all I wrote above is completely correct?
2) Could you please write me down step by step number II. ? 
Thank you!
 A: Be careful: it is true that $P^e=\{x/s:x\in P,s\in S\}$, but it's not automatic that $x/s\in P^e$ implies $x\in P$, as you seem to be using. This is true for prime ideals not intersecting $S$. Indeed, if
$$
\frac{x}{s}=\frac{y}{t}
$$
with $y\in P$, then $u(tx-sy)=0$, for some $u\in S$; therefore $tux=suy\in P$; since $tu\notin P$ (as $P\cap S=\emptyset$), we conclude $x\in P$. But notice that the statement doesn't generally hold for arbitrary ideals of $R$.
Let's state it officially.

Lemma. If $P$ is a prime ideal of $R$ and $P\cap S=\emptyset$, then $x/s\in P^e$ if and only if $x\in P$.

With the above lemma, we can shorten the proofs.
Proof of Ia
Let $P$ be a prime ideal of $R$, with $P\cap S=\emptyset$. If $P^e=S^{-1}R$, then $1/1\in P^e$, hence $1\in P$: contradiction.
Suppose now that $(x/s)(y/t)\in P^e$; then $(xy)/(st)\in P^e$, so $xy\in P$.
Proof of Ib
If $f\colon A\to B$ is a ring homomorphism and $P$ is a prime ideal of $B$, then $f^{-1}(P)$ is a prime ideal of $A$. This follows immediately from the definitions. In particular, if $P'$ is a prime ideal of $S^{-1}R$, then $(P')^c=\nu^{-1}(P')$ is a prime ideal of $R$. If $s\in (P')^c\cap S$, then $s/1\in P'$ by definition, so $P'$ contains an invertible element: contradiction.
Proof of II
Consider a prime ideal $Q$ of $S^{-1}R$ and set $P=Q^c$. Let's show that $P^e=Q$. Suppose $x/s\in Q$. Then also
$$
\nu(x)=\frac{x}{1}=\frac{x}{s}\frac{s}{1}\in Q
$$
Therefore $x\in P=Q^c$ and therefore $x/s\in P^e$.
Suppose $x\in P$ and take any $s\in S$. Then $\nu(x)=x/1\in Q$ and so
$$
\frac{x}{s}=\frac{x}{1}\frac{1}{s}\in Q
$$
Conversely, if $Q=P^e$, for some prime ideal $P$ of $R$, then $P=Q^c$ (easy proof).
