We have that a property of the transition operator defining c-equivalence (or star equivalence from equation 1 in Bertelson) is
\begin{align*} T(f\star_Mg)=T(f)\star'T(g)\,, \end{align*}
where $\star_M$ is the Moyal star product and $\star'$ is a star product for another ordering. Therefore, the stargenvalue equation, $H\star_MW=EW$, becomes, upon operation under the transition operator,
\begin{align*} T(H\star_MW)=T(EW)\implies T(H)\star'T(W)=ET(W)\,, \end{align*} where $H$ is the Hamiltonian, $W$ is the Wigner function, and $E$ is the energy spectrum. We note that $E$ does not change under the transition operator.
We can derive $H\star_MW=EW$ using the star exponential (defined in Section 4 in Bayen),
\begin{align*} Exp_{\star M}\left(\frac{tH}{i\hbar}\right)=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{t}{i\hbar}\right)^nH^{\star_Mn}\,, \end{align*}
where $t$ is the time and
\begin{align*} H^{\star_Mn}=\underbrace{H\star_MH\star_M\cdots\star_MH}_{\text{$n$ times}} \end{align*}
When switching to a different ordering using different star products, I've noticed that several papers (for example, Section 3.3 in Dito and Section 5 in Hirshfeld) write the star exponential as
\begin{align*} Exp_{\star'}\left(\frac{tH}{i\hbar}\right)=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{t}{i\hbar}\right)^nH^{\star'n}\,, \end{align*}
without using the transition operator. Why is this allowed?
I'm really confused about this because a different energy spectrum can then be calculated, which contradicts $T(H)\star'T(W)=ET(W)$.
$\Large{\text{Update}}$
Looking at the published version by D&T, I was able to keep track of the $\exp(\gamma t/2)$ term. So, I see how (pg. 313)
\begin{align*} T\left(Exp_{\star M}\left(\frac{tH}{i\hbar}\right)\right)=Exp_{\star\gamma}\left(\frac{tT(H)}{i\hbar}\right) \end{align*}
compares to
\begin{align*} Exp_{\star\gamma}\left(\frac{tH}{i\hbar}\right)(q,p)=\exp(\gamma t/2)\sum_{n=0}^{\infty}\big(-i(n+1/2)\omega t\big)T(\pi_n) \end{align*}
According to equation (10),
\begin{align*} Exp_{\star}\left(\frac{tH}{i\hbar}\right)=\sum_{n=0}^{\infty}e^{\frac{tE_n}{i\hbar}}\pi_n\,. \end{align*}
So, if $\star=\star_M$,
\begin{align*} T\left(Exp_{\star M}\left(\frac{tH}{i\hbar}\right)\right)&=T\left(\sum_{n=0}^{\infty}e^{\frac{tE_n}{i\hbar}}\pi_n\right)\\ &=\sum_{n=0}^{\infty}e^{\frac{tE_n}{i\hbar}}T(\pi_n)\,, \end{align*}
As
\begin{align*} Exp_{\star\gamma}\left(\frac{tT(H)}{i\hbar}\right)=Exp_{\star\gamma}\left(\frac{tH}{i\hbar}\right)\exp(-\gamma t/2)\,, \end{align*}
by using
\begin{align*} T\left(Exp_{\star M}\left(\frac{tH}{i\hbar}\right)\right)=Exp_{\star\gamma}\left(\frac{tT(H)}{i\hbar}\right)\,, \end{align*}
wouldn't that mean,
\begin{align*} T\left(Exp_{\star M}\left(\frac{tH}{i\hbar}\right)\right)=\sum_{n=0}^{\infty}\big(-i(n+1/2)\omega t\big)T(\pi_n)\,, \end{align*}
so that $E_n=(n+1/2)\hbar\omega$? However, if we use equation (10) with $\star=\star_{\gamma}$, then I immediately see how to get $E_n=\hbar\omega(n+1/2+i\gamma/2\omega$. Both ways seem valid to me, so I must be missing a crucial detail.