I think it should work as follows.
Let $\Omega$ be a vector cyclic and separating for $N$. Then $$\Omega_2=\begin{pmatrix} \Omega & 0\\0 &\Omega \end{pmatrix}$$ is cyclic and separating for $M=N\otimes \mathbb M_2(\mathbb C)$, where $M$ acts by left multiplication. If we can choose isometries $s,t\in N$ satsifying $ss^\ast +tt^\ast =1$, with $s^\ast s =t^\ast t=1$ and $t^\ast s =s^\ast t =0$ then $$\Phi= \begin{pmatrix} s^2\Omega &st\Omega \\ ts\Omega & t^2\Omega\end{pmatrix} =v\Omega_2\,, \qquad v=\begin{pmatrix} s^2 &st \\ ts & t^2\end{pmatrix} $$ should be a joint cyclic and separating vector for $N\otimes I_2$ and $M$. Because we can write every element in $M$ as $$ \begin{pmatrix} a &b \\c & d\end{pmatrix} = \begin{pmatrix} as^\ast s^\ast+b t^\ast s^\ast + c s^\ast t^\ast +d t^\ast t^\ast& 0\\0& as^\ast s^\ast+b t^\ast s^\ast + c s^\ast t^\ast +d t^\ast t^\ast\end{pmatrix} \begin{pmatrix} s^2 &st \\ ts & t^2\end{pmatrix}\,,$$ we have $ (N\otimes I_2)v = M $ and in particular $\Phi$ is cyclic for $N\otimes I_2$. Further $\Phi$ is cyclic for $M'$ because $\Omega_2$ is cyclic for $M'$ and $\Phi=v\Omega_2$ with $v\in M$.
In the case asked, $N=B(H)$ and $H$ separable, let $\{e_0,e_1,e_2,\ldots\}$ be a ONB of $H$. Define e.g. isometries by $se_i=e_{2i}$ and $te_i=e_{2i+1}$ in $B(H)$. Take $B(H)$ in the standard form acting on the Hilbert-Schmidt operators $L^2(H)$. Then any positive element $\Omega\in L^2_+(H)$ will be cyclic and separating for $B(H)$.