If $A$ is a commutative ring with unit, $a \in A $ and $S = \lbrace a^n \mid n \geq 0 \rbrace $ then there is an isomorphism $$S^{-1}A \cong A[x]/(1-ax).$$
In fact we can consider the homomorphism $$\phi : A[x] \to S^{-1}A$$ $$p(x) \mapsto p\left(\frac{1}{a} \right)$$
Then obviously $(1-ax) \subseteq \ker(\phi) $, but why $\ker(\phi) \subseteq (1-ax) $ ?
You can take that path, but another one is easier: show that $A[x]/(1-ax)$ has the universal property of $S^{-1}A$, using the universal property of the polynomial ring $A[x]$ and of quotients.
Spoiler
That is, suppose we have a map $\eta:A\to B$ where $\eta a$ is invertible. Then by the universal property of $A[x]$; there is a unique morphism $\bar \eta:A[x]\to B$ such that $x\mapsto (\eta a)^{-1}$ and $\bar\eta\iota=\eta$, $\iota:A\to A[x]$ the inclusion. Then clearly $1-ax\in \ker \bar \eta $, so $(1-ax)\subseteq\ker \bar\eta$. By the universal property of the quotient there is a unique morphism $\tilde{\eta}:\tilde A=A[x]/(1-ax)\to B$ such that $\tilde\eta\pi =\bar\eta$, whence $\tilde\eta(\pi\iota)=\bar\eta\iota=\eta$. Finally if $\tilde \iota=\pi\iota$ $\tilde\iota(a)=a+(1-ax)$ has inverse $x+(1-ax)$; so every element of $S$ is mapped to a unit by $\tilde\iota$. It follows $(\tilde A,\tilde\iota)$ satisfies the universal property of $(S^{-1}A,\iota_S)$ so these are isomorphic.
You can find the calculation in Voloch's note Rings of fractions the hard way. Here is another proof which uses the universal properties: If $B \in \mathsf{CRing}$, we have natural bijections
$$\hom(A[x]/(1-ax),B) \cong \{g \in \hom(A[x],B) : g(1-ax)=0\}$$ $$ = \{(f,b) \in \hom(A,B) \times |B| : f(a) b = 1\} \cong \{f \in \hom(A,B) : f(a) \in B^*\} \cong \hom(A[a^{-1}],B).$$
Hence, Yoneda tells us $A[x]/(1-ax) \cong A[a^{-1}]$.
Let $p(x)$ of degree $n$ be such that $ p(\frac{1}{a})=0$ in $S^{-1}A$ this means that for some $k$, $ a^kp(\frac{1}{a})=0$ in $A$ where we take $k\geq n$. In other words $ x^kp(\frac{1}{x})$ is a polynomial in $A[x]$ with a zero at $a$. Further the degree of $ x^kp(\frac{1}{x})$ is at most $k$. (depending on the lowest nonzero term in $p(x)$)
This means that there is $q(x)$ of degree at most $k-1$, such that
$$ x^kp(\frac{1}{x})=(x-a)q(x)$$ Now this means we also have $$ \left(\frac{1}{x}\right)^kp(x)=(\frac{1}{x}-a)q(\frac{1}{x})$$ and on multiplying by $x^k$ we have $$p(x)=(1-ax)x^{k-1}q(\frac{1}{x})$$ now $x^{k-1}q(\frac{1}{x})$ is a polynomial in $A[x]$ since the degree of $g$ is at most $k-1$. m Therefore $p(x) \in (1-ax)$.
