Let $M$ be a von Neumann algebra and $\rho$ is a normal state on $M$. If there exist von Neumann algebras $M_1$ and $M_2$ such that $M\cong M_1\bar{\otimes} M_2$.
Can we conclude that there exist normal states $\rho_i$ on $M_i(i=1,2)$ such that $\rho=\rho_1\otimes\rho_2$?
No, absolutely not. For a simple example, take $M_1=M_2=M_2(\mathbb C)$ and $\rho$ the Bell state $$ \rho=\frac 1{\sqrt 2}(E_{11}\otimes E_{11}+E_{22}\otimes E_{22}). $$ If it were of the form $\rho=\rho_1\otimes\rho_2$, then we would have $\rho_1=\mathrm{tr}_2(\rho)$, $\rho_2=\mathrm{tr}_1(\rho)$, i.e., $$ \rho_1=\rho_2=\frac 1{\sqrt 2}(E_{11}+E_{22}). $$ But clearly $\rho_1\otimes \rho_2\neq \rho$.
I'll leave it as an exercise to show that this is even wrong for finite-dimensional commutative von Neumann algebras $M_1$ and $M_2$.