Let $A$ be an ultraweakly dense sub-$C^*$-algebra of a $W^*$-algebra $M$.
Is it possible to represent $M(A)$, the two-sided multiplier algebra as a sub-$C^*$-algebra of $M$ so that the inclusion $A\subset M$ is respected?
In other words, do we really have to take $A^{**}$ in the definition of the multiplier algebra if already $A$ sits nicely in some $W^*$-algebra?
There is a natural bijective correspondence between the elements of $M(A)$ and the multipliers of $A$ in $M$. In other words, you always have $$ M(A)=\{x\in A'':\ xA\subset A,\ Ax\subset A\} $$ for any faithful representation of $A$.
First note that any $\rho\in M(A)$ is bounded as a left multiplication operator $\rho:A\to A$. Indeed, if you suppose otherwise there would exist elements $z_n\in A$ with $\|z_n\|<\frac1n$ and $\|\rho(z_n)\|>n$. If we put $a=\sum_nz_nz_n^*$, using a bit of functional calculus it can be shown that since $z_nz_n^*\leq a$ then there exists $v_n\in A$ with $\|v_n\|\leq\|a^{1/6}\|$ and $z_n=a^{1/3}v_n$. This gives $$ \|\rho(z_n)\|=\|\rho(a^{1/3}v_n)\|=\|\rho(a^{1/3})v_n\|\leq\|\rho(a^{1/3})\|\,\|v_n\|\leq \|\rho(a^{1/3})\|\,\|a^{1/6}\|, $$ a contradiction.
Fix an approximate unit $\{e_j\}$ in $A$. Then $\{\rho(e_j)\}$ is (by the above) a bounded net in $M$, so it has a wot cluster point $x\in M$. For any $y\in A$ we have that $\{e_jy\}$ converges in norm to $y$, so $\rho(e_jy)\to\rho(y)$. We have the equality $\rho(e_jy)=\rho(e_j)y$, and taking wot limit we get $\rho(y)=xy$. So $xy\in A$ and $x$ is a left multiplier. If $z\in M$ satisfies $zy=\rho(y)$ for all $y\in A$, then $(x-z)y=0$ for all $y\in A$. As $A$ is dense in $M$, we get that $(x-z)(x-z)^*=0$, and so $z=x$. So each multiplier corresponds to a unique element in $M$.
If we see $\rho$ as a right multiplier we can repeat the above, and in principle we obtain (as a left multiplier) $x_1\in M$ such that $x_1y=\rho(y)$ and (as a right multiplier) $yx_2=\rho(y)$ for all $y\in A$. Then for $y\in A$ we have $$ yx_2=\rho(y)=x_1y. $$ As we can choose for $y$ a net that converges to $1\in M$ we obtain that $x_2=x_1$.
Conversely, given $x\in M$ such that $xa,ax\in A$ for all $a\in A$, we can repeat the previous argument but using now the inclusion $A\subset A^{**}$ to get $\rho\in A^{**}$ corresponding to $x$. Hence the correspondence $\rho\leftrightarrow x$ is a bijection. The fact that this is a bijection both as left and right multipliers, immediately implies that this is a $*$-isomorphism between the two multiplier algebras.