Here I will follow Blackadar's book (PDF link to extended verison). Theorem III.2.2.8 here says the following. Let $M,N$ be von Neumann algebras acting on Hilbert spaces $H_M, H_N$ respectively, and let $\pi:M\rightarrow N$ be a (normal) $*$-isomorphism. Then:
$\pi$ is equivalent a reduction of an amplification of the identity representation of $M$. This is a standard result, in lots of textbooks, and I completely understand the proof. To be precise, it says there is $K$ and a projection $e' \in M' \bar\otimes \mathcal B(K)$, and a unitary $u:H_N \rightarrow e'(H_M\otimes K)$ such that $\pi(x) = u^* ( e'(x\otimes 1) ) u$. (So here $e'$ is the "reduction" restricting $M\otimes 1_K$ to an invariant subspace).
I now quote: "Thus there is a Hilbert space $H'$ such that $M\otimes 1$ on $H_M\otimes H'$ is unitarily equivalent to $N\otimes 1$ on $H_N\otimes H'$."
The second part is not proved, and I do not see how to show it. I've seen similiar statements in the literature, but I cannot find this statement in other textbooks. Can anyone indicate a proof, or does anyone have a reference with a proof?
The closest I can find to a reference in a book is in Bratteli, Robinson, Operator algebras and quantum statistical mechanics vol 1. Theorem 2.4.26 which just says "Cantor-Bernstein" (which as I'll explain shortly, is rather natural) without further detail. Thanks to MaoWao for the link to Jones's lecture notes, which essentially gives the details, by utilising the Comparison Theory of Projections (the key result which we need here is indeed a Cantor-Bernstein like result, which I should have remembered, but didn't of course).
The "standard" part of the result quoted in the questions shows, in particular, that for suitable $K$ there is an isometry $j: H_N \rightarrow H_M\otimes K$ with $(x\otimes 1)j = j\pi(x)$ for $x\in M$. Clearly we may replace $K$ by any Hilbert space of equal or larger dimension, and for a suitably large $K$ we have $K\otimes K\cong K$, and so we obtain an isometry $u:H_N\otimes K \rightarrow H_M\otimes K\otimes K \cong H_M\otimes K$ with $(x\otimes 1)u = u(\pi(x)\otimes 1)$ for $x\in M$.
Now apply the same argument to $\pi^{-1}$, again enlarging $K$ if necessary, to obtain an isometry $v:H_M\otimes K \rightarrow H_N\otimes K$ with $(\pi(x)\otimes 1)v = v(x\otimes 1)$ for $x\in M$.
So we have isometries going both ways intertwining the dilated representations, and hence "Cantor-Bernstein" should build us a unitary. To actually prove this, let's follow Jones and use comparison theory.
Consider $M$ acting on $(H_M\otimes K)\oplus(H_N\otimes K)$ as $$ x\mapsto \begin{pmatrix} x\otimes 1 & 0 \\ 0 & \pi(x)\otimes 1 \end{pmatrix}. $$ Then $(x\otimes 1)u=u(\pi(x)\otimes 1)$ precisely means that $$ u' = \begin{pmatrix} 0 & u \\ 0 & 0 \end{pmatrix} \in M'. $$ Note that $$ (u')^* u' = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \quad u' (u')^* = \begin{pmatrix} uu^* & 0 \\ 0 & 0 \end{pmatrix} \leq \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, $$ which follows as $uu^* \leq 1$ because $u$ is an isometry. This means that $$ \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \preceq \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}. $$ Exactly the same argument with $v$ shows that $$ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \preceq \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}. $$ Now consult your favourite textbook which will show that $\preceq$ is a partial order which in particular means that $$ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \approx \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} $$ (The proof of this is, when you think about it, exactly an analogue of the Cantor-Bernstein result.) This means there is a partial isometry $w\in M'$ with $$ w^*w = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad ww^* = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} $$ But this precisesly means that $w$ restricts to a unitary map $H_M\otimes K \rightarrow H_N\otimes K$ and $w\in M'$ means that this unitary is an intertwiner, as required.