This is far from a full answer (but maybe it will inspire other answers).

I think all the cases Yemon gives, we only get a cb map, but we don't get normality. However, in some special cases, you do get a normal map.

Firstly, if $M=L^\infty(G)$ with $\Delta(f)(s,t) = f(st)$, then define $T_*:L^1(G) \rightarrow L^1(G\times G)$ by
$$T_*(f)(s,t) = f(st) f_0(s),$$
where $f_0\in L^1(G)$ is some fixed, positive function with $\|f_0\|_1=1$. This is bounded, as
\begin{align} \|T_*(f)\|_1 &= \int_G \int_G |f(st) f_0(s)| \ dt \ ds
= \int_G \int_G |f(ss^{-1}t)| \ dt \ |f_0(s)| \ ds \\
&= \int_G \|f\|_1 |f_0(s)| \ ds = \|f\|_1, \end{align}
using the left-invariance of the Haar measure.
Then the pre-adjoint $\Delta_*:L^1(G\times G)\rightarrow L^1(G)$ is
$$\Delta_*(f)(t) = \int_G f(s,s^{-1}t) \ ds,$$
i.e. convolution (if you set $f=a\otimes b$). So then
$$\Delta_* T_*(f) (t) = \int_G T_*(f)(s,s^{-1}t) \ ds
= \int_G f(ss^{-1}t) f_0(s) \ ds = f(t).$$
So $\Delta_* T_*$ is the identity, as required.

More generally, and as Yemon alludes to in a comment, if $(M,\Delta)$ is a compact Kac algebra, then it's operator biprojective (see Aristov's paper "Amenability and compact type for Hopf-von Neumann algebras from the homological point of view"). So we can choose $T$ to be cb and normal (and with various $M_*$ module properties). If we can choose $T$ with these module properties (but not necessarily normal) then $M$ is said to be operator biflat. For $VN(G)$ this was investigated by Aristov, Runde and Spronk, "Operator biflatness of the Fourier algebra and approximate indicators for subgroups". It seems to be unknown if $VN(G)$ ever fails to be operator biflat. Of course, this is a much stronger condition that Yemon asks for.

It seems surprising to me that we can just write down a suitable (normal) map $T$ for $L^\infty(G)$, but that it seems that Yemon's question is open for $VN(G)$.

**Edit (29 Nov):** (This is technical; I hope I got all the details correct). Suppose $G$ is a locally compact group such that $VN(G)$ admits a faithful normal trace (I think is true if $G$ is a separable SIN group, for example). Then let $\varphi$ be the Plancheral weight on $VN(G)$; let $\omega$ be a normal tracial state. Then $\psi=\varphi\otimes\omega$ is a semifinite trace on $VN(G\times G)=VN(G)\overline\otimes VN(G)$, and for $x\in VN(G)_+$, we have that $$ \psi(\Delta(x)) = \varphi((\iota\otimes\omega)\Delta(x)) = \omega(1) \varphi(x) = \varphi(x) $$ as $\varphi$ is invariant for $\Delta$ (I know this from the quantum group setting; see papers of Kustermans and Vaes). Let $(\sigma_t)$ be the modular automorphism group for $\varphi$; then $(\sigma_t\otimes\iota)\Delta(x) = \Delta(\sigma_t(x))$ for $x\in VN(G)$, and so as $\omega$ is a trace, we see that $\Delta(VN(G))$ is invariant for the modular automorphism group of $\psi$. By a theorem of Takesaki there is a normal conditional expectation $\epsilon:VN(G\times G)\rightarrow \Delta(VN(G))$ with $\psi(x) = \psi(\epsilon(x))$ for all $x$ in the definition ideal of $\psi$.

So in particular, this gives a positive answer (even with the "normal") condition in the separable SIN case, say.