There is at least a case when it _is_ true though.  Suppose $G$ itself is compact (this argument doesn't need $G$ abelian), so that $bG = G$.  The $M(G)$ is the _multiplier algebra_ of $L^1(G)$ and as $L^1(G)$ has a contractive approximate identity, there is an isometric embedding $M(bG) = M(G) = M(L^1(G)) \rightarrow L^1(G)^{**}$.

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Let me sketch this.  Let $A$ be a Banach algebra with contractive approximate identity $(e_\alpha)$.  I will regard the multiplier algebra $M(A)$ as double centralisers: pairs of maps $L,R$ from $A$ to $A$ with
$$ L(ab) = L(a)b, \qquad R(ab) = aR(b), \qquad aL(b) = R(a)b \qquad (a,b\in A). $$
It turns out that, using the approximate identity, one can show that $L,R$ are automatically linear, and also (closed graph theorem) that $L,R$ are bounded.  (Or make this part of the definition, if you wish).

Turn $A^*$ and $A^{**}$ into $A$-bimodules in the usual way.  Given $(L,R)\in M(A)$ let $x^{**}\in A^{**}$ be an accumulation point of the bounded net $(L(e_\alpha))$.  For $x^*\in A^*$ and $x\in A$ compute:
$$ \langle x^{**} \cdot a, x^* \rangle = \langle x^{**}, a \cdot x^* \rangle
= \lim_\alpha \langle a \cdot x^*, L(e_\alpha) \rangle
= \lim_\alpha \langle x^*, L(e_\alpha)a \rangle
= \lim_\alpha \langle x^*, L(e_\alpha a) \rangle
= \langle x^*, L(a) \rangle. $$
Thus $x^{**}\cdot a = L(a)$ (or the canonical image thereof in $A^{**}$).  Similarly,
$$ \langle a \cdot x^{**}, x^* \rangle
= \lim_\alpha \langle x^*, a L(e_\alpha) \rangle
= \lim_\alpha \langle x^*, R(a) e_\alpha \rangle
= \langle x^*, R(a) \rangle, $$
so that $a\cdot x^{**} = R(a)$.  This gives us the required embedding.

For those that know about Arens products, there are clear links.  I believe this construction is due to McKilligan ([MathSciNet](https://mathscinet.ams.org/mathscinet-getitem?mr=318881) or [JLMS Article](https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/jlms/s2-6.3.399)).

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As Nik Weaver notices, of course $C(bG) = C(G) \subseteq L^\infty(G)$ and so we obtain a quotient map $\theta : L^\infty(G)^{*} \rightarrow M(G)$.  Let $\phi:M(G)\rightarrow L^1(G)^{**}$ be the map we just constructed.  Let $\mu\in M(G)$ so the associated double centraliser is $L(f)  =\mu * f, R(f) = f * \mu$ for $f\in L^1(G)$.  Then for $F\in C(G)$ (and writing $\cdot$ for the module actions, which are related to but not quite equal to convolution),
$$ \langle \theta(\phi(\mu)), F \rangle = \langle \phi(\mu), F \rangle_{L^\infty(G)^*, C(G)}. $$
Now, a bounded approximate identity argument and a calculation shows that every $F\in C(G)$ is equal to $f\cdot F'$ for some $F'\in C(G)$ and $f\in L^1(G)$.  Thus
$$ \langle \theta(\phi(\mu)), F \rangle = \langle F', \mu * f \rangle_{C(G), L^1(G)}
= \ip{\mu}{F}. $$
So $\theta \circ \phi$ is the identity, and hence $M(bG)$ is a complemented subspace of $L^\infty(G)^*$ in this case.

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I have had a quick think, and I cannot see how to say much in the non-compact case.