Let $S$ be a multiplicative subset of a commutative ring $R$ with identity and let $I$ be an ideal of $R$. Then $S^{-1}I= S^{-1}R$ if and only if $S\cap I\neq \emptyset$.
Proof: If $s\in S\cap I$, then $1_{s^{-1}R}=s/s \in S^{-1}I$ and hence $S^{-1}I = S^{-1}R$. Conversely, if $S^{-1}I = S^{-1}R$, then $\varphi_S^{-1}(S^{-1}I) = R$ whence $\varphi_S(1_R) = a/s$ for some $a\in I$, $s\in S$. Since $\varphi_S(1_R) = 1_Rs/s$ we have $s^2s_1 = ass_1$ for some $s_1 \in S$. But $s^2s_1 \in S$ and $ass_1 \in I$ imply $S\cap I\neq \emptyset$.
The map $\varphi_S:R\to S^{-1}R$ is given by $\varphi_S(r)=rs/s$ (for any $s\in S$) and $S^{-1}I=\{a/s\mid a\in I; s\in S\}$.
Question: I don’t understand the point of second sentence in the proof. To be specific, use of map $\varphi_S$ and $R$ has a multiplicative identity. Suppose $S^{-1}I=S^{-1}R$. We know $S^{-1}R$ is a commutative ring with multiplicative identity $s/s$, where $s\in S$. By hypothesis, $s/s \in S^{-1}I$. So $\exists a\in I$ such that $s/s=a/s \Leftrightarrow s^2s_1=ass_1$. We got to our desired result. Why we need to assume $R$ has a identity?
You do not. Hungerford often works in this section with rings with identity, though this is not strictly necessary. Hungerford does this mostly for convenience; some authors define multiplicative subsets to be submonoids of the multiplicative monoid of $R$, which would force multiplicative subsets to contain $1$ (and necessarily be nonempty), just as some authors require rings to have an identity. You should not be surprised if one of the theorems in this section includes an assumption that $R$ has an identity but the result holds without that assumption as well.
Second, you need to reorganize the argument you present: it first appears that you select an $s\in S$, and then describe elements of $S^{-1}I$ in the form $\frac{a}{s}$ with that $s$ you previously selected. That is not necessarily possible: you need not be able to express an element of $S^{-1}I$ with a pre-specified denominator.
So instead it should proceed as follows: if $S^{-1}I=S^{-1}R$, then let $s\in S$. Then $\frac{s}{s}\in S^{-1}R = S^{-1}I$, so there exists $a\in I$ and $t\in S$ such that $\frac{a}{t}=\frac{s}{s}$. Therefore, there exists $s'\in S$ such that $s'(sa-st)=0$, so $s'sa = s'st$. Since $s'sa\in I$ and $s'st\in S$, it follows that $I\cap S\neq\varnothing$.
Hungerford's argument is perfectly fine, so I do not see why you worry about "the point". The point is that it proves what Hungerford wants to prove. That he does so under hypotheses that are not as restrictive as possible does not detract from their correctness.
Why would he want to structure it like this? Because this will be the structure of the argument in Lemma 4.9 and other places, so he is introducing the technique he will use later.