All rings are unital and commutative. Here are parts a. and b. of the exercise from Eisenbud. I don't think that part a. is needed for b., but I've included it just in case.
Suppose $R$ is a $\mathbb{Z}$-graded ring and $0 \neq f \in R_1$. Show that $R[f^{-1}]$ is again a $\mathbb{Z}$-graded ring. Let $S = R[f^{-1}]_0$.
a. Show that $R[f^{-1}] \cong S[x,x^{-1}]$, where $x$ is a new variable
b. Show that $S = R[f^{-1}]_0 \cong R/(f-1)$.
I am able to prove the first statement, where I defined the degree $n$ component of $R[f^{-1}]$ to be $$ R[f^{-1}]_n = \left\{ \frac{r}{f^d} : r \ \text{is homogeneous}, \deg(r) - d = n \right\}, $$ and I found part a. to be straightforward.
I am able to prove part b. when $f$ is not a zerodivisor as follows: define $\phi: R \to S$, by $r \mapsto \frac{r}{f^{\deg(r)}}$ for homogeneous $r$, and then extending to all of $R$ by linearity. It is easy to see the $\phi$ is onto, so it suffices to show that $\ker \phi = (f-1)$. The inclusion $(f-1) \subseteq \ker \phi$ is obvious. On the other hand, suppose that $r = r_1 + \cdots r_n \in \ker \phi$, where each $r_i$ is homogeneous of degree $d_i$. Moreover, assume that $d_1 < \cdots < d_n$. Then $$ \phi(r) = \frac{r_1}{f^{d_1}} + \cdots + \frac{r_n}{f^{d_n}} = \frac{f^{d_n-d_1}r_1 + \cdots + f^{d_n-d_{n-1}}r_{n-1} + r_n}{f^{d_n}} = 0, $$ which implies $f^{d_n-d_1}r_1 + \cdots + f^{d_n-d_{n-1}}r_{n-1} + r_n = 0$, since I am assuming that $k$ is not a zerodivisor (otherwise I would have to multiply this equation by some $f^k$, which would screw up the next step). But then $$ r = r_1 + \cdots + r_n = (r_1 + \cdots + r_{n-1}) - (f^{d_n-d_1}r_1 + \cdots + f^{d_n-d_{n-1}}r_{n-1}) = (1 - f^{d_n - d_1}) r_1 + \cdots + (1 - f^{d_n - d_{n-1}}) r_{n-1}, $$ which is in $(f-1)$ as desired.
The claim is also true in the case where $f$ is nilpotent: then the localization at $f$ is $0$, and $1-f$ is a unit, implying that $R/(f-1) = 0$.
So really I'm after the case where $f$ is a zerodivisor, but not nilpotent.
In general, when $\phi(r)=0$, it means that there is $k \geq 0$ such that $f^k(f^{d_n-d_1}r_1+\ldots+r_n)=0$. Then $r=(1-f^k)r+ f^kr=(1-f^k)r+f^k(r_1+\ldots+r_{n-1})-f^k(f^{d_n-d_1}r_1+\ldots+f^{d_n-d_{n-1}}r_{n-1})=(1-f^k)r+f^k\sum_{i=1}^{n-1}{(1-f^{d_n-d_i})r_i}$ and each term is in $(f-1)$, so $r \in (f-1)$.
Edit: here’s a somewhat simpler solution, probably easier to apply in general. Let $r \in R$ be in the kernel of $\phi$. It’s easy, writing $r$ in coordinates and “completing” the terms of lower degree with powers of $f$, to show that $r=(1-f)r_1+r’$ for some $r_1,r’ \in R$ and $r’$ homogeneous of degree $d$. Thus $\phi(r’)=0$. But $\Phi(r’)=\frac{r’}{f^d}$, so this means that there is a $k \geq 1$ such that $f^kr’=0$. But then $r’=(1-f^k)r’ \in (1-f)$.