Let $M \subseteq B(H)$ be a von Neumann algebra. Given $S \subseteq M$, let $VNA(S)$ be the von Neumann algebra generated by $S$. Note that I require von Neumann algebras to be unital, so $VNA(S) = S''$ (I think).
If $v \in VNA(x)$ where $x \in M$, do we also have $v^*v \in VNA(x^*x)$?
Attempt: $v$ is a ultraweak limit of polynomials in $x$ and $x^*$, but I can't conclude from this that $v^*v$ is a limit of polynomials in $x^*x$.
The von Neumann algebra generated by a set is $(S\cup S^*)''$.
Anyway, the answer to your question is no. Take $M=M_2(\mathbb C)$, and $$ x=\begin{bmatrix} 0&1\\1&0\end{bmatrix}, \qquad v=\begin{bmatrix} 1&1\\1&1\end{bmatrix} . $$ Then $v\in W^*(x)$. In fact, $v\in\operatorname{span}\{x\}$, since $v=I+x$. But $x^*x=1$, so $W^*(x^*x)=\mathbb C\,1$, while $v^*v=2v$ is not a scalar multiple of the identity.