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
$$ \langle \theta(\phi(\mu)), F \rangle = \langle F', \mu * f \rangle_{C(G), L^1(G)}
= \langle \mu, F \rangle. $$
So $\theta \circ \phi$ is the identity, and hence $M(bG)$ is a complemented subspace of $L^\infty(G)^*$ in this case.
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I have had a quick think, and I cannot see how to say much in the non-compact case.
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**Edit:** Some comments on uniqueness, prompted by interesting questions of Nik Weaver. Given a bounded approximate identity $(e_\alpha)$ let $x_0^{**} \in A^{**}$ be some accumulation point. Then our embedding $M(A)\rightarrow A^{**}$ is $(L,R) \mapsto L^{**}(x^{**}_0)$, which is isometric if $\|x^{**}_0\|=1$. Notice that if $x^{**} = L^{**}(x^{**}_0)$ then $x^{**}\cdot a, a\cdot x^{**} \in A \subseteq A^{**}$ for each $a\in A$.
Conversely, if $x^{**}\in A^{**}$ is any element with $x^{**}\cdot a, a\cdot x^{**} \in A$ for each $a\in A$, then we can define linear maps $L,R:A\rightarrow A$ with $L(a) = x^{**}\cdot a$ etc. and then $(L,R)\in M(A)$. It is _tempting, but wrong_ to think that we have shown that
$$ M(A) \cong \{ x^{**}\in A^{**} : A\cdot x^{**}, x^{**}\cdot A \subseteq A \}. $$
What goes wrong is that we can have a non-zero $x^{**}\in A^{**}$ with $A\cdot x^{**} + x^{**}\cdot A =\{0\}$. In the example of $A=L^1(G)$ we know that $A\cdot A* + A^*\cdot A$ is the sum of the left/right uniformly continuous functions. So any $x^{**}\in L^\infty(G)^*$ which annihilates these, but is non-zero, induces the zero multiplier. Notice that this cannot happen for $C^*$-algebras for example, as here $A^*\cdot A = A\cdot A^* = A^*$.
So, the embedding of $M(A)$ into $A^{**}$ depends on the choice of $x_0^{**}$ an accumulation point of a bai of $A$. You can characterise such $x_0^{**}$ as the "mixed identities" of $A^{**}$ (a right identity for the 1st Arens product and a left identity for the 2nd Arens product). Any such mixed identity is some accumulation point of some bai of $A$.