Let $R$ be a ring and let $S \subset R$ be a multiplicative system, i.e. for all $f,g \in S$ it follows $fg \in S$. Let $S^{-1}R$ denote the localization of $R$ that is defined using the relation on $R \times S$ \begin{align*}(a,\alpha) ∼(b,\beta) \;\Leftrightarrow\; \exists s \in S: s\cdot(a\beta-b\alpha)=0.\end{align*} As $S^{-1}R$ is a ring again, it is possible to localize this ring again. According to my lecture notes there should be a canonical isomorphism $S^{-1}S^{-1}R \cong S^{-1}R$. However, there is no construction provided in these notes.
I am aware of the fact that the localization can be defined using universal properties but I am not quite familiar with the theory of categories. Therefore, I want to ask if somebody can provide this isomorphism for me and maybe provide a proof without using category theory.
As @Lukas Heger has suggested we need to proof that the natural ring morphism \begin{align*} \phi:A \rightarrow S^{-1}A, \enspace\;a \mapsto \frac{a}{1} \end{align*} is an isomorphism if $S \subset A^*$. For this we can use the property $\ker(\phi) = \{a \in A\mid \exists t\in S: t\cdot a = 0\}$, which proves the injectivity of $\phi$. Also, since $S \subset A^*$, we can represent any class $\frac{b}{s}\in S^{-1}A$ as $\frac{s^{-1}b}{1}$. Thus this morphisms is also subjective and also an isomorphism.
Now, it is be definition that $S \subset (S^{-1}A)^*$. Therefore, this previous statement proves the existence of an isomorphism $S^{-1}S^{-1}A \cong S^{-1}A$.