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 $\{s_1, \dots, s_n\}$.