Let $\mathfrak{a}$ be an ideal of a ring $R$, and let $S = 1 + \mathfrak{a}$. Show that $S^{-1}\mathfrak{a}$ is contained in the Jacobson radical of $S^{-1}R$.
We note that every elemement in $S$ is of the form $1+a$ for some $a \in \mathfrak{a}$. Therefore any element in $\mathbb{a} \subset R$ is mapped to $\frac{a}{1+a'}$ for some $a' \in \mathfrak{a}$, via the map $f: R \to S^{-1}R$. To show that $S^{-1}\mathfrak{a}$ is contained in the Jacobson radical of $S^{-1}R$, it suffices to show that for each $a,a' \in \mathfrak{a}$ and for all $r \in R$, $1 - \frac{ar}{1+a'}$ is a unit in $S^{-1}R$.
We note that $1+a \in S$ and therefore $1+a$ is a unit in $S^{-1}R$. It follows that $a$ is nilpotent and conseqentially $\frac{ar}{1+a'}$ is nilpotent. We therefore conclude that $1 - \frac{ar}{1+a'}$ is a unit in $S^{-1}R$.
Can I make the argument any more rigorous?
The map $f\colon R\rightarrow S^{-1}R$ is $r\mapsto \frac r1$. Hence $a\in \mathfrak a$ is mapped to $\frac a1$ under $f$ (i. e. $a'=0$ in your notation).
From the fact that $1+a$ is invertible in $S^{-1}R$ it does not follow that $a$ is nilpotent (e.g. look at $R = k[X]$ and $\mathfrak a = (X)$. Then $S^{-1}R\subseteq k(X)$ and $1+X$ is invertible, but $X$ is not nilpotent).
For any $a\in \mathfrak a$ and $r\in R$ you have $1 - \frac{ra}1 = \frac{1-ra}{1}$ and as the denumerator clearly lies in $1+\mathfrak a = S$, it has to be invertible.