Errors have been fixed
In this second response let me try to address what might play the role of the operator sum $\tilde{J_1} + \cdots + \tilde{J_n}$ which is the analogue of $\sum_{t \in T} \, t \ \in \Bbb{C}[S_n]$ where $T$ is the conjugacy class consisting of all transpositions.
Given two sets of generic parameters $x_1, \dots, x_{n-1}$ and and $y_1, \dots, y_{n-2}$ the associated Okada algebra $\mathcal{F}_n$ has a presentation given by generators $E_1, \dots, E_{n-1}$ subject to the defining relations
\begin{equation} \begin{array}{rll} E_i^2 &= \ x_i \, E_i & \\ E_i \, E_j &= \ E_j \, E_i &\text{whenever $|i-j| >1$} \\ E_j \, E_i \, E_j &= \ y_i \, E_j &\text{whenever $j = i+1$} \end{array} \end{equation}
Given a word $w$ the Okada power-symmetric functions $p_w$ is
defined by the recursion
\begin{equation} p_w \ = \ \left\{ \begin{array}{ll} \displaystyle x_1 \cdots \, x_k &\text{if $\, w = 1^k \, $ with $\, k \geq 0$} \\ \displaystyle q_k \cdot p_u \big[ + (k + 2) \big] &\text{if $\, w= u 2 1^k \,$ with $\, k \geq 0$} \end{array} \right\} \end{equation}
where
\begin{equation} q_k \ := \ (x_1 \cdots \, x_k) \, \Big( x_{k+1} x_{k+2} \, - \, (k+2) \,y_{k+1} \Big) \end{equation}
and $p_u \big[ +l \big]$ means perform the substitutions $x_i \mapsto x_{i+l}$ and $y_i \mapsto y_{i+l}$ for all $i \geq 1$.
For the purposes of this discussion, let's provisionally define the Okada power-symmetric function in terms of the Okada-Schur functions and the Okada characters values:
\begin{equation} p_u \, = \ \sum_{|v| = |u|} \, X^v_u \, s_v \end{equation}
Now let's define $\mathcal{F_n}$-valued versions $\Bbb{s}_w$ and $\Bbb{p}_w$ of the Okada-Schur and power-symmetric functions. For $n \geq k \geq 1$ define the following $k \times k$ tri-diagonal determinants whose values are in the Okada algebra $\mathcal{F_n}$
\begin{equation} \begin{array}{ll} \mathcal{P}_k \, := &\det \, \begin{pmatrix} x_1 & x_2 E_1 & 0 & \cdots \\ 1 & x_2 & x_3 E_2 & & \\ 0 & 1 & x_3 & & \\ \vdots & & & \ddots \end{pmatrix} \\ \\ \\ \mathcal{Q}_{k-1} \, := &\det \, \begin{pmatrix} x_2 E_1 & x_1 x_3 E_2 & 0 & \cdots \\ 1 & x_3 & x_4 E_3 & & \\ 0 & 1 & x_4 & & \\ \vdots & & & \ddots \end{pmatrix} \end{array} \end{equation}
where $\mathcal{P}_0 := 1$.
Clearly an order must be observed
when tabulating the determinant --- following the conventions of
Kerov and Goodman, the $l$-th factor in the expansion will always
be selected from the $l$-th column.
The values of the determinants $\mathcal{P}_k$ and $\mathcal{Q}_{k-1}$ are in fact independent of the order in which products are taken in the Laplace expansion: This is because indices of the generators $E_1, \dots, E_{n-1}$ which participate in any given monomial in the expansion of such a tri-diagonal determinant will always differ by at least two.
Employ the same recursion above being mindful to place the accumulating $\mathcal{Q}$-factors to the left and in order:
\begin{equation} \Bbb{s}_w \, := \ \left\{ \begin{array}{ll} \mathcal{P}_k &\text{if $w= \, 1^k \, $ and $k \geq 0$} \\ \\ \mathcal{Q}_k \, \big[+ |v| \big] \cdot \Bbb{s}_v &\text{if $w= \, 1^k \, 2 \, v \, $ and $k \geq 0$} \end{array} \right\} \end{equation}
Once again $\mathcal{Q}_k \big[ + |v| \big]$ means shift all indices by $|v|$. Play the same game and define the $\mathcal{F}_n$-valued power-symmetric functions by the expansion:
\begin{equation} \Bbb{p}_u \, := \ \sum_{|v| = |u|} \, X^v_u \, \Bbb{s}_v \end{equation}
I want to make use of Okada's trace functional $\text{Tr}: \mathcal{F}_n \longrightarrow \Bbb{C}$ which is defined for an element $a \in \mathcal{F}_n$ using the Okada-Schur values by
\begin{equation} \text{Tr}(a) \ = \ {1 \over {(x_1 \cdots \, x_n)}} \, \sum_{|v|=n} \, s_v \cdot \text{tr} \Big[ \sigma_v(a) \Big] \end{equation}
where $\sigma_v : \mathcal{F}_n \longrightarrow \text{End}(V_v)$ is the irreducible representation of $\mathcal{F}_n$ associated to a word $v$ with $|v|=n$ and where $\text{tr} \big[ \cdot \big]$ denotes the usual trace. Okada proves in his paper that
\begin{equation} \begin{array}{ll} \displaystyle \text{Tr} \, (1) &\displaystyle = \, 1 \\ \displaystyle \text{Tr} \, \big( ab \big) &\displaystyle = \, \text{Tr} \, \big(ba \big) \\ \displaystyle \text{Tr} \, \big( aE_i \big) &\displaystyle = \, {y_i \over {x_{i+1}}} \, \text{Tr} \, (a) \quad \text{when $a \in \mathcal{F}_i$} \quad \left( { \scriptstyle \begin{array}{l} \text{As far as I can tell there seems to be} \\ \text{a missing $x_{i+1}$ in the denominator of} \\ \text{part (4) of Proposition 2.7 in Okada's} \\ \text{paper which I have tried to correct here.} \end{array} }\right) \end{array} \end{equation}
Using these multiplicative properties, a simple induction on the length $|v|$ reveals that $\text{Tr} \, \big( \Bbb{s}_v \big) \, = \, s_v$ and consequently $\text{Tr} \, \big( \Bbb{p}_u \big) \, = \, p_u$. Moreover
\begin{equation} \begin{array}{c} \displaystyle \text{tr} \, \Big[ \pi_v \big( \Bbb{p}_u \big) \Big] \, = \, X^v_u \\ \displaystyle \text{--- and ---} \\ \displaystyle \text{tr} \, \Big[ \pi_v \big( \Bbb{s}_u \big) \Big] \, = \, \delta_{u,v} \end{array} \end{equation}\begin{equation} \begin{array}{c} \displaystyle \text{tr} \, \Big[ \pi_v \big( \Bbb{p}_u \big) \Big] \, = \, \big( x_1 \cdots x_n \big) \, X^v_u \\ \displaystyle \text{--- and ---} \\ \displaystyle \text{tr} \, \Big[ \pi_v \big( \Bbb{s}_u \big) \Big] \, = \, \big(x_1 \cdots x_n \big) \, \delta_{u,v} \end{array} \end{equation}
Isn't this observation an indication that $\Bbb{p}_u$ is an analogue of a characteristic function of a conjugacy class in the group setting ? However, pursuing this analogy, it's not immediately clear whether the elements $\mathcal{p}_u$ are central in $\mathcal{F}_n$.
yours, Ines.
p.s. In fact the order of products taken in the expansion of the $\mathcal{F}_n$-valued determinants $\mathcal{P}_k$ and $\mathcal{Q}_{k-1}$ is irrelevant --- this is because all monomials which occur involve $E$-generators whose subscripts differ by at least two.