The proof of (LC3), in the given setting, is surprisingly difficult, or, at least, elaborate. Let $a \in A$ with $\mu_S(a) = [a] = 0$ in $A[S^{-1}]$, i.e. one has \begin{equation} \tag{1} a \in (sT_s-1\,|\,s \in S). \end{equation} To show is that \begin{equation} \tag{2} sa = 0 \end{equation} for some $s \in S$. Because of (1), there are elements $s_1, \dots s_n \in S$ and polynomials $p_1(T), \dots,p_m(T) \in A[T]$ such that \begin{equation} a = \sum_{i=1}^n p_i(T) (s_iT_i - 1) \quad \text{in $A[T]$},\quad,\quad T_i := T_{s_i}. \end{equation} As a first reduction, we may assume $p_i(T) = p_i(T_1, \dots,T_n)$ for all $i$, so that \begin{equation} \tag{3} a = \sum_{i=1}^n p_i(T_1, \dots, T_n) (s_iT_i - 1) \quad \text{in $A[T]$}. \end{equation} Namely, let $T' \subseteq T$ be those indeterminates which either equal some $T_i$, or which appear in at least one $p_i(T)$, $i = 1, \dots, n$, so that we may write $T' = \{T_1, \dots, T_n, T_{n+1}, \dots, T_q\}$. By eventually introducing dummy terms with coefficient 0, we may assume $p_i(T) = p_i(T') = p_i(T_1, \dots, T_q)$, so that $$a = \sum_{i=1}^n p_i(T_1, \dots, T_q) (s_iT_i - 1)$$. Putting $p_i(T):=0$ for $i=n+1, \dots, q$ then gives \begin{equation*} a = \sum_{i=1}^q p_i(T_1, \dots, T_q) (s_iT_i - 1) \quad \text{in $A[T]$}, \end{equation*} which upon renaming $q$ by $n$ gives (3).
To prove that $sa = 0$ for some $s \in S$ we proceed by induction on $n$. For $n = 1$ we start with
\begin{equation*}
a = p(T_s) (sT_s - 1) \quad \text{in $A[T]$}
\end{equation*}
for some indeterminate $T_s \in X$. We abbreviate notation by writing $u := T_s$, so that we have the equation
\begin{equation}
a = p(u) (su - 1) \quad \text{in $A[T]$}.
\end{equation}
Let $p(u) = \sum_{k=0}^d a_k u^k$; then
\begin{equation*}
\begin{split}
p(u) (su - 1)
&= \sum_{k=0}^d sa_k u^{k+1}-\sum_{k=0}^d a_ku^k\\
&= \sum_{k=1}^{d+1} sa_{k-1} u^k -
\sum_{k=0}^d a_k u^k\\
&= sa_du^d + \sum_{k=1}^d(sa_{k-1}-a_k) u^k-a_0\\
&= a,
\end{split}
\end{equation*}
so that
\begin{equation*}
a_0=-a \quad,\quad a_k=sa_{k-1}\,,\,k=1,\dots, d-1
\quad,\quad sa_d = 0,
\end{equation*}
hence
\begin{equation*}
a_k = -s^ka \,,\, k=0, \dots, d \quad,\quad
sa_d = 0 ,
\end{equation*}
so that
\begin{equation*}
s^{d+1}a = -sa_d = 0,
\end{equation*}
as was to be shown. This establishes the base clause of the induction.
We now assume that $n \ge 1$, and that, with $k < n$, \begin{equation*} a = \sum_{i=1}^k p_i(T_1,\dots,T_n)(s_iT_i-1) \quad \text{in $A[T]$} \end{equation*} implies that $sa = 0$ for some $s \in S$, and we want to show that the same is true for $k = n$. So we assume, with a given ring $A$, that $a \in \ker \mu_S$ and (2) holds. We put $A' := A[T_n]/(s_nT_n - 1)$. The projection $A \longrightarrow A'$ then realizes(!) the localization $$\mu_{S'} : A \longrightarrow A[S'^{-1}]$$ with $S' := \{s_n\}$; in particular, $A'= A[S'^{-1}]$. The canonical map \begin{equation*} A[T_n] \longrightarrow A[T] \longrightarrow A[S^{-1}] \end{equation*} induces, by passing to the quotient, $$A'= A[S'^{-1}] \longrightarrow A[S^{-1}] = (A[S'^{-1}])[S^{-1}]$$, which realizes the localization \begin{equation*} \mu_S' : A[S'^{-1}] \longrightarrow (A[S'^{-1}]) [S^{-1}]. \end{equation*} The localization map $\mu_S : A \longrightarrow A[S^{-1}]$ then factors as the composite of localizations \begin{equation*} A \longrightarrow A' \longrightarrow A[S^{-1}] = A \longrightarrow A[S'^{-1}] \longrightarrow (A[S'^{-1}])[S^{-1}]. \end{equation*} Let $\overline{a} \in A' = A[S'^{-1}]$ be the image of $a \in A$ under $A \longrightarrow A'$. Then $\mu_S(a) = \mu_S'(\overline{a}) = 0$. and so, by (3), \begin{equation*} \overline{a} = \sum_{i=1}^{n-1} \overline{p_i}(T_1, \dots, T_{n-1}) (s_iT_i - 1) \quad \text{in $A'[T]$} \end{equation*} with $\overline{p_i}(T_1, \dots, T_{n-1}) = p_i(T_1,\dots, T_{n-1},1/s_n)$, $i=1, \dots, n-1$, since $s_nT_n - 1 = 0$ in $A' = A[S'^{-1}]$. Therefore, by the induction hypothesis, $s\overline{a} = \overline{sa} = 0$ for some $s \in S$. Thus $sa \in \ker \mu_{S'}$, and so, by the base clause $n=1$ applied to $\mu_{S'}$, \begin{equation*} s_n^{d+1}(sa) = (s_n^{d+1}s)a = 0, \end{equation*} which finishes the proof. As a byproduct of the proof we obtain that $s$ in (2) may be chosen as a product of the $s_i$'s (with repeated factors), i.e. as an element of the multiplicative closure of $\{s_1, \dots, s_n\}$ .