Let $I = \langle \{sX_s - 1 : s \in U\}\rangle.$ I'm looking for an alternate proof of $U^{-1}R \cong R[\{X_s\}_{s \in U}]/I.$
The proof I found is sending $r/u \to rX_u,$ proving this is a well-defined homomorphism, then giving the inverse and checking it's an inverse. However, proving it's a well-defined homomorphism involves many more steps than you would expect, none of the properties are given for free.
The approach everyone practically leaps out of their chairs to suggest when seeing the problem is sending $r \to r/1, X_s \to 1/s,$ then proving the kernel is $I.$ Everything is well-defined and nothing additional has to be checked.
Certainly $I \subseteq \ker \phi,$ but the other direction appears impossible. For example, if $0 = \phi(aX_u+b) = a/u + b,$ we get $v(a+bu) = 0$ for some $v \in U.$ We would like to prove $a+bu = 0$ so that $aX_u + b = -b(uX_u - 1) \in I,$ but $v$ might be a zero divisor, so this is not guaranteed. $vt = 0$ for $t \in U$ would make $U^{-1}R$ trivial so that there is nothing to show, but perhaps $t \in R \setminus U.$ What then?
As the last paragraph suggests, we can show that for all $x \in \ker \phi,$ there exists $u \in U$ such that $ux \in I.$ So for any $x \in \ker \phi / I,$ we have $x = (uX_u)x = (ux)X_u \in I \Rightarrow x = 0$ for some $u.$ The result now follows from $R[\dots]/\ker \phi \cong (R[\dots]/I)/(\ker \phi / I).$