Elementary symmetric Jucys-Murphy evaluations are cycle-count class sums

Base case. For $n=1$ the list $X_2,\dots ,X_n$ is empty, so $e_0=1$ by [F2], and the right-hand side for $s=0$ is the single class sum $C_{(1)}^{(1)}$ of the identity of $S_1$, whose cycle type $(1)$ has $\mathrm{\ell }=1=n-0$; for $s\ge 1$ the left-hand side is $e_s$ of no variables, hence $0$ by [F2], and no partition $\rho ⊢1$ has $\mathrm{\ell }(\rho )=1-s\le 0$. Thus the identity holds for $n=1$ and all $s$.

Induction hypothesis. Let $n\ge 2$ and assume that for all $0\le s\le n-1$ one has $e_s(X_2,\dots ,X_{n-1})={\sum }_{\rho ⊢n-1,\mathrm{\ell }(\rho )=n-1-s}{C}_{\rho }^{(n-1)}$ in $\mathbb{Z}[S_{n-1}]$, the sum being $0$ when no such partition exists.

Recursion for elementary symmetric polynomials. For $m\ge 1$ and $0\le s\le m$, splitting the $s$-element subsets of $\{1,\dots ,m\}$ into those not containing $m$ and those containing $m$ gives $e_s(x_1,\dots ,x_m)=e_s(x_1,\dots ,x_{m-1})+x_m{e}_{s-1}(x_1,\dots ,x_{m-1})$ in any commutative ring, with the convention $e_{-1}=0$; the identity is trivial for $s=0$ as well.

First summand. For $0\le s\le n-1$, take $m=n-1$ and $x_i=X_{i+1}$ in step 1.3, using [F5] and use [F1] to identify $X_k$ in $\mathbb{Z}[S_{n-1}]$ with $X_k$ in $\mathbb{Z}[S_n]$: $e_s(X_2,\dots ,X_{n-1})={\sum }_{\rho ⊢n-1,\mathrm{\ell }(\rho )=n-1-s}{C}_{\rho }^{(n-1)}$ by step 1.2. Each class sum $C_{\rho }^{(n-1)}$ is the sum of the permutations $\sigma \in S_{n-1}$ of cycle type $\rho$; regarded in $S_n$ such a $\sigma$ fixes $n$ and has one further cycle, namely $(n)$, so its cycle type in $S_n$ has $\mathrm{\ell }(\rho )+1=n-s$ cycles; conversely every $\tau \in S_n$ with $\tau (n)=n$ and $n-s$ cycles restricts to a permutation of $S_{n-1}$ with cycle type $\rho$ and $\mathrm{\ell }(\rho )=n-1-s$. Hence this summand equals the sum of all $\tau \in S_n$ with $\tau (n)=n$ and exactly $n-s$ cycles.

Second summand. For $s=0$ this summand is zero by $e_{-1}=0$. For $1\le s\le n-1$, with the same substitution, $X_n{e}_{s-1}(X_2,\dots ,X_{n-1})=({\sum }_{j<n}(jn))e_{s-1}(X_2,\dots ,X_{n-1})={\sum }_{j<n}{\sum }_{\sigma }(jn)\sigma$, where $\sigma$ runs over the permutations of $S_{n-1}$ with exactly $n-s$ cycles and the second identity uses step 1.2 for $s-1$ and [F1]. For such a $\sigma$ and $j$, use [F4] to write $\sigma$ as a product of pairwise disjoint cycles and insert the fixed point $j$ as a $1$-cycle if necessary; if $(j{c}_1\dots c_b)$ is the cycle of $j$, then evaluating on the letters $j,c_1,\dots ,c_b,n$ shows $(jn)\sigma$ replaces that factor by the single cycle $(nj{c}_1\dots c_b)$ and keep all other factors, so $(jn)\sigma$ moves $n$ and has the same number $n-s$ of cycles as $\sigma$. The map $(\sigma ,j)\mapsto \tau :=(jn)\sigma$ is a bijection from these pairs onto the permutations $\tau \in S_n$ with $\tau (n)\mathrm{\ne }n$ and $n-s$ cycles, with inverse $\tau \mapsto ((\tau (n)n)\tau ,\tau (n))$: indeed $j:=\tau (n)$ lies in $\{1,\dots ,n-1\}$, the product $(jn)\tau$ fixes $n$ and hence lies in $S_{n-1}$, and the two constructions invert one another because $(jn{)}^2=1$. Hence this summand is the sum of all $\tau \in S_n$ with $\tau (n)\mathrm{\ne }n$ and exactly $n-s$ cycles, each occurring once.

Adding the two summands via step 1.3, every $\tau \in S_n$ with exactly $n-s$ cycles is counted exactly once, according to whether $\tau (n)=n$ or $\tau (n)\mathrm{\ne }n$; grouping by cycle type and using [F3] gives $e_s(X_2,\dots ,X_n)={\sum }_{\rho ⊢n,\mathrm{\ell }(\rho )=n-s}{C}_{\rho }^{(n)}$. For $s=0$ this reads $1=C_{(1^n)}^{(n)}$; for $s=1$ it sums the permutations with $n-1$ cycles, exactly the transpositions when $n\ge 2$; when $n=1$ both the transposition sum and $e_1$ vanish; for $s=n-1$ it sums the permutations with a single cycle, the $n$-cycles. For $s>n-1$ the left-hand side is $e_s$ in the $n-1$ variables $X_2,\dots ,X_n$, hence $0$ by [F2], and the right-hand side is an empty sum.