= Solution
The dense commutant orbit makes $\Omega$ separating for $M$, so $S_0(a\Omega)=a^*\Omega$ is well defined on the dense subspace $M\Omega$. Put $\tau(a)=\langle a\Omega,\Omega\rangle$. The assumed identity says that this positive <vector> functional is a <tracial positive functional>. Hence
$$
\|S_0(a\Omega)\|^2=\tau(aa^*)=\tau(a^*a)=\|a\Omega\|^2.
$$
It extends to an antiunitary involution $J$ on $H$, with $J\Omega=\Omega$. This is the <modular conjugation>, and the <modular operator> in this tracial case is $\Delta=I$.
On the dense subspace $M\Omega$,
$$
(JaJ)(b\Omega)=J(ab^*\Omega)=ba^*\Omega.
$$
Thus it acts by right multiplication and commutes with all left multiplications; $JMJ\subseteq M'$. We prove the reverse inclusion rather than assuming the general Tomita commutation theorem.
Let $T\in M'$, $C=\|T\|$, and $\eta=T\Omega$. The set
$$
\mathcal B=\{a\Omega:a\in M,\ \|a\|\leq C\}
$$
is <weakly compact> and convex. Indeed, <trace-class duality> and the <Banach-Alaoglu theorem> make the bounded-operator ball compact in a topology stronger than the <weak operator topology>. The algebra $M$ is weak-operator closed, and $a\mapsto a\Omega$ is weakly continuous, so its image is <weakly compact>.
For $b\in M$ use the <polar decomposition of a bounded operator> $b=uh$, $h=|b|$. Both $h$ and $u$ belong to $M$: <continuous functional calculus> gives $h$, and $b(h+\varepsilon I)^{-1}$ tends strongly to $u$. The polar decomposition is proved independently in Question 8. Since $T$ commutes with $M$,
$$
\langle T\Omega,b\Omega\rangle
=\langle T h^{1/2}u^*\Omega,h^{1/2}\Omega\rangle.
$$
The <trace> identity gives
$$
\|h^{1/2}u^*\Omega\|^2=\tau(uhu^*)=\tau(hu^*u)=\tau(h),
\qquad \|h^{1/2}\Omega\|^2=\tau(h).
$$
Therefore $|\langle\eta,b\Omega\rangle|\leq C\tau(h)$. Meanwhile
$$
\sup_{a\in M,\ \|a\|\leq C}\operatorname{Re}\langle a\Omega,b\Omega\rangle
=C\tau(h),
$$
with the supremum attained by $a=Cu$. For the upper bound, traciality gives $\tau(b^*a)=\tau(h^{1/2}u^*a h^{1/2})$; evaluating this on $\Omega$ bounds its absolute value by $\|a\|\tau(h)$.
These support inequalities imply $\eta\in\mathcal B$. Otherwise the <Hahn-Banach separation theorem> would separate $\eta$ from this closed <convex set> by a real inner-product functional. Since $M\Omega$ is dense, its separating <vector> could be approximated by some $b\Omega$, preserving strict separation, contrary to the displayed inequality. Thus $T\Omega=a\Omega$ for an $a\in M$ of <norm> at most $C$.
The <bounded operator> $R_a=Ja^*J$ commutes with $M$ and sends $\Omega$ to $a\Omega$. It agrees with $T$ on the cyclic <vector> and therefore on all of $M\Omega$, hence on $H$. This proves $T\in JMJ$, and so
$$
\boxed{JMJ=M'.}
$$
For the regular representation, take $\Omega=\delta_e$. It is cyclic for the left action and for the commuting right action. On finite sums of left translations, its <vector> functional is the identity coefficient, so $\tau(ab)=\tau(ba)$. This extends to the whole <group von Neumann algebra> by the bounded strong-star approximations of Question 6, taking limits of the <vector> functional and products. The conjugation is explicitly
$$
(J\xi)(h)=\overline{\xi(h^{-1})},\qquad J\lambda(g)J=\rho(g).
$$
Consequently
$$
\boxed{\lambda(\Gamma)'=J\lambda(\Gamma)''J=\rho(\Gamma)''.}
$$
Thus the commutant in (b) is exactly the <Von Neumann algebra> of right regular translations, with the inverse in $\rho(g)$ fixing the representation convention.
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