= Solution
The relevant <centralizer of a subalgebra> is
$$
Z_{(n-1,1)}
=\{a\in\mathbb C S_n:ab=ba\text{ for all }b\in\mathbb C S_{n-1}\}.
$$
Equivalently, it is the fixed subspace of $\mathbb C S_n$ under conjugation by $S_{n-1}$. We prove the <Olshanskii centralizer lemma> directly, without assuming <multiplicity-free restriction>.
A permutation has one cycle containing $n$, of some length $r\geq1$. Let $\nu$ record the lengths of its other nontrivial cycles, omitting fixed points. Two permutations are conjugate under $S_{n-1}$ exactly when these data $(r,\nu)$ agree: a relabelling of the other letters identifies the distinguished cycles and all remaining cycles. Hence the sums $K_{r,\nu}$ over these conjugation orbits form a basis of $Z_{(n-1,1)}$.
Filter these orbit sums by the number of moved letters other than $n$:
$$
t=(r-1)+|\nu|.
$$
If $r=1$, the permutation fixes $n$, and $K_{1,\nu}$ is already a <conjugacy-class sum> in $Z_{n-1}$. For $r\geq2$, let $C_\nu\in Z_{n-1}$ be the class sum with nontrivial cycle lengths $\nu$. Expand
$$
X_n^{\,r-1}C_\nu.
$$
Terms in which the $r-1$ star <transpositions> use distinct letters, disjoint from the support of the chosen term of $C_\nu$, produce precisely $K_{r,\nu}$, each with coefficient one. Indeed, a cycle through $n$ has a unique expression as an ordered product of star <transpositions> using each of its other letters once; the remaining cycles uniquely determine the term of $C_\nu$.
Every other term moves fewer than $t$ letters outside $n$. A repeated letter uses at most $r-2$ distinct such letters in the star product, and an overlap with the support of $C_\nu$ likewise reduces the size of their union. Since the whole product commutes with $S_{n-1}$, these remaining terms are a <linear combination> of lower-filtered orbit sums. Thus
$$
X_n^{\,r-1}C_\nu=K_{r,\nu}+\sum_{(r',\nu'):\ r'-1+|\nu'|<t}c_{r',\nu'}K_{r',\nu'}.
$$
Induction on $t$, starting with the identity, puts every orbit sum in $\operatorname{alg}(Z_{n-1},X_n)$. The reverse inclusion holds because both proposed generators commute with $\mathbb C S_{n-1}$. Therefore
$$
\boxed{Z_{(n-1,1)}=\operatorname{alg}(Z_{n-1},X_n)}.
$$
In particular this <centralizer of a subalgebra> is commutative: $Z_{n-1}$ is commutative and $X_n$ commutes with it.
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