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 $M(bG) = M(G) = M(L^1(G)) \rightarrow L^1(G)^{**}$.
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 or JLMS Article).
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 $$ \langle \theta(\phi(\mu)), F \rangle = \langle F', \mu * f \rangle_{C(G), L^1(G)} = \langle \mu, F \rangle. $$ So $\theta \circ \phi$ is the identity, and hence $M(bG)$ is a complemented subspace of $L^\infty(G)^*$ in this case.
I have had a quick think, and I cannot see how to say much in the non-compact case.