Let $T\in B(\mathcal{H} \otimes \mathcal{H})$ where $\mathcal{H}$ is a Hilbert space. We can define operators $$T_{[12]}= T \otimes 1;\quad T_{[23]}= 1 \otimes T$$ and if $\Sigma: \mathcal{H} \otimes \mathcal{H} \to \mathcal{H} \otimes \mathcal{H}$ is the "flip" map, then we can define $$T_{[13]}= \Sigma_{[23]}T_{[12]}\Sigma_{[23]}= \Sigma_{[12]}T_{[23]}\Sigma_{[12]}$$

Question: Given $S,T \in B(\mathcal{H} \otimes \mathcal{H})$, is it true that $$(ST)_{[13]}= S_{[13]}T_{[13]}?$$

I attempted this as follows:

We know that the algebraic tensor product $B(\mathcal{H}) \odot B(\mathcal{H})$ is weak$^*$-dense (= $\sigma$-weakly dense) in $B(\mathcal{H} \otimes \mathcal{H})$. It is easy to see that the identity holds for $S,T \in B(\mathcal{H}) \odot B(\mathcal{H})$.

Can I conclude from this that the equality holds for all $S,T \in B(\mathcal{H}) \overline{\otimes} B(\mathcal{H})= B(\mathcal{H}\otimes \mathcal{H})$ (here, the first tensorproduct is the von-Neumann algebraic tensor product).

It is natural to try to use results involving weak$^*$-continuity and Kaplansky-density-like results, but I'm having trouble finishing the proof. Any ideas?