For a commutative ring $R$ with $1\neq 0$ and a nonzerodivisor $r \in R$, let $S$ be the set $S=\{r^n\mid n\in \mathbb{Z}, n\geq 0\}$ and denote $S^{-1}R=R\left[\frac{1}{r}\right]$. Prove that there is a ring isomorphism $$R\left[\frac{1}{r}\right]\cong \frac{R[x]}{(rx -1)}.$$
I'm thinking maybe I can find a homomorphism from $R[x]$ to $R\left[\frac{1}{r}\right]$ that has kernel $(rx-1)$, and then use the first isomorphism theorem. Is this the right approach?
On
Tricky is proving the kernel $K = (rx\!-\!1).\,$ A simple way: if $f\in K$ then by nonmonic division
$$ r^n f(x) = (rx\!-\!1)\,q(x) + r',\ \ {\rm for} \ \ r'\in R,\ n\in \Bbb N$$
Evaluating at $\, x = 1/r\,$ shows $\,r'\! = 0\,$ so $\,rx\!-\!1\mid r^n f\,\Rightarrow\,rx\!-\!1\mid f,\,$ by $\,(rx\!-\!1,r) = (1);\,$ more explicitly $\,rx\!-\!1\mid rg\,\Rightarrow\, rx\!-\!1\mid g = x(rg)-(rx\!-\!1)g$.
Remark $\ $ See this answer for another proof and further discussion. If you already know basic (universal) properties of localizations then see the linked dupe for ways to employ them.
You should note that $\frac{1}{r}$ has the property $\frac{1}{r}\cdot r=1$. That is, $\frac{1}{r}$ is not just an indeterminate object like $x$. Conceptually, this isomorphism should be easy to construct. Send $s\in R$ to itself for all $s$. Send $x\mapsto \frac{1}{r}$. This is clearly a surjective homomorphism $$ R[x]\xrightarrow{\phi} S^{-1}R.$$ Calculate $\ker\phi$. You will see that it is $(rx-1)$. Conclude using the first isomorphism theorem.