Given a Hilbert space $\mathcal{H}$.
Consider a von Neumann algebra: $$M\subseteq\mathcal{B}(\mathcal{H}):\quad M=M''$$
Suppose a cyclic vector: $$\Omega\in\mathcal{H}:\quad\overline{\mathcal{M}\Omega}=\mathcal{H}$$
Regard the involution: $$S_0:\mathcal{M}\Omega\to\mathcal{H}:\quad S_0M\Omega:=M^*\Omega$$ This operator is closable!
But I didn't succeed checking this?
What if there's no cyclic vector?
The arguments I know require that $\Omega$ is both cyclic and separating (i.e., also cyclic for the commutant).
When you only require that $\Omega$ is separating, I don't think that $S_0$ even makes sense. For instance, let $\mathcal M=B(\mathcal H)$, $\Omega=e_1$. Consider the (constant) sequence $E_{12}e_1=0$. Then $S_0E_{12}e_1=E_{21}e_1=e_2\ne0$, so $S_0$ is not even well-defined.
Now, if $\Omega$ is cyclic and separating, the map $S_0$ is well-defined, because if $M\Omega=0$, then $M=0$ by the separating property.
The key to the proof is to also consider the related map $F_0:\mathcal M'\Omega\ni N\Omega\longmapsto N^*\Omega\in \mathcal H$.
Note also that $S_0$ and $F_0$ are conjugate-linear (so some inner products below will look like they are the wrong way).
We have, for $M\in\mathcal M$ and $N\in\mathcal M'$, $$ \langle S_0M\Omega,N\Omega\rangle= \langle M^*\Omega,N\Omega\rangle=\langle N^*M^*\Omega,\Omega\rangle=\langle N^*\Omega,M\Omega\rangle. $$ This implies that $N\Omega$ is in the domain of $S_0^*$ and that $S_0^*N\Omega=N^*\Omega$. So $S_0^*$ is a (closed) extension of $F_0$. Then $S_0^*$ is densely defined, and it follows that $S_0$ is closable.