Let $G$ be a discrete group and $M = L G \subset B(\ell^2 G)$ its left regular von Neumann algebra. According to [Section 13.1.3, AP] it is possible to associate to any given (unitary) group representation $\pi \in \mathrm{Rep}(G)$ a $L G$-bimodule, ie a Hilbert space $\mathcal{H}(\pi)$ with two normal, commuting left and right representations of $L G$, by fixing $\mathcal{H}(\pi) = \ell^2(G) \otimes H_\pi$, where $\pi : G \to B(H_\pi)$, with the left and right actions given by extension of $$ \lambda_g \cdot ( \delta_k \otimes \xi) \cdot \lambda_h = \delta_{g \, k \, h} \otimes \pi(g)\xi. $$ Observe that by Fell's absorption principle, the left representation is unitarity equivalent to $1 \otimes \lambda$ and so, it extends to the von Neumann algebra $L G$. According to [Section 13.1.3, p. 216, AP]:
Obviously, the equivalence class of $\mathcal{H}(\pi)$ only depends on the equivalence class of the representation $\pi$. The trivial $L G$-$L G$ -bimodule $L^2(M)$ is associated with the trivial representation of $G$ and the coarse bimodule $L^2(M) \otimes L^2(M)$ corresponds to the left regular representation (Exercise 13.12) [emphasis mine]
I have some questions filling out the details of the bold part.
In the text above, the equivalence of representation is unitary equivalence, while two bimodules $\mathcal{H}$ and $\mathcal{K}$ are isomorphic iff there is a unitary isometry $U: \mathcal{H} \to \mathcal{K}$ that is also bimodular, ie: $U(a \cdot \xi \cdot b ) = a \cdot U(\xi) \cdot b$. It is clear that $\mathcal{\pi_1} \cong \mathcal{\pi_2}$ implies that $\mathcal{H}(\pi_1) \cong \mathcal{H}(\pi_2)$, since a unitary isometry $U: H_{\pi_1} \to H_{\pi_2}$ intertwining the representations gives a unitary $id \otimes U: \mathcal{H}(\pi_1) \to \mathcal{H}(\pi_2)$ that is bimodular.
But the reciprocal seems a bit more contentious to me. Let $\pi_1$, $\pi_2$ be two representations and assume for simplicity they act on the same space $H$. Let $T: \ell^2(G) \otimes H \to \ell^2(G) \otimes H$ be a bounded linear map. We can represent it as a matrix indexed by $G \times G$ and with entries in $\mathcal{B}(H)$, $[T_{g,h}]_{g,h \in G}$. Now we have: \begin{eqnarray} T \mbox{ is right } M\mbox{-modular } & \Longleftrightarrow & \big[T_{g \, k, \, h \, k}\big]_{g,h} = \big[ T_{g,h} \big]_{g,h}, \quad \forall k \in G.\\ T \mbox{ is left } M\mbox{-modular } & \Longleftrightarrow & \big[T_{k \, g, \, k \, h}\big]_{g,h} = \big[ \pi_2(k) \, T_{g,h} \, \pi(k)^{-1} \big]_{g,h}, \quad \forall k \in G. \end{eqnarray} The first equation implies that $T$ is of the from $\sum_{g \in G} T_{g} \otimes \lambda_g$, where $T_{h \, g^{-1}} = T_{g,h}$. Which combined with the second equation gives that $$T_{k \, g \, k^{-1}} = \pi_1(k) \, T_{g} \, \pi_2(k^{-1})$$. This makes clear the fact that if $T_e \neq 0$, $T_e$ is an intertwining operator. That is like saying that, after identifying the operators satisfying the first equation with a subset of $L G \bar\otimes \mathcal{B}(H)$ that condition is equivalent to having $(id \otimes \tau)(T) \neq 0$. But this of course doesn't follow automatically. Indeed, if $G$ is not icc and $\pi_1 = \pi_2 = 1$, then $\mathcal{H}(1) = \ell^2(G)$ and any unitary $U$ inside the center $\mathcal{Z}(L G)$ is an intertwining map such that $U_e = \tau(U)$, but we can take the last term $= 0$.
Q: Why is the reciprocal $\mathbf{\mathcal{H}(\pi_1) \cong \mathcal{H}(\pi_2) \Rightarrow \pi_1 \cong \pi_2}$ true? Am I missing something?, it doesn't look ''obvious'' to me.