$Y$ can be described as the Stone space of the measure algebra of $(X,\mu)$. That is, let $\Sigma$ be the $\sigma$-algebra on which $\mu$ is defined, let $N\subseteq\Sigma$ be the ideal of null sets, and let $B=\Sigma/N$ be the quotient Boolean algebra. Then $Y$ is naturally homeomorphic to the set $S$ of Boolean homomorphisms $B\to\{0,1\}$, topologized as a subspace of $\{0,1\}^B$.
To prove this, let us first recall that $Y$ can be described as the set of $*$-homomorphisms $L^\infty(X,\mu)\to\mathbb{C}$, with the topology of pointwise convergence. For each $b\in B$, there is a function $1_b\in L^\infty(X,\mu)$, and a $*$-homomorphism $\alpha$ must send $1_b$ to either $0$ or $1$ since $1_b^2=1_b$. It is then easy to see that $b\mapsto \alpha(1_b)$ is a Boolean homomorphism $B\to\{0,1\}$ (the Boolean operations on sets can be expressed in terms of ring operations on their characteristic functions). This defines a map $F:Y\to S$.
Note moreover that since simple functions are dense in $L^\infty(X,\mu)$, an element of $Y$ is determined by its values on characteristic functions $1_b$. Thus $F$ is injective. Also, $F$ is continuous, since the topology on $S$ is the topology of pointwise continuity with respect to evaluation at just the elements $1_b$. Since $Y$ and $S$ are both compact Hausdorff, it follows that $F$ is an embedding.
It remains to be shown that $F$ is surjective. Fix a homomorphism $h:B\to\{0,1\}$, and let $U=h^{-1}(\{1\})$. The idea is that we can then define a $*$-homomorphism $L^\infty(X,\mu)\to\mathbb{C}$ which maps a function $f$ to the "limit" of the values of $f$ along the ultrafilter $U$. To make this precise, given $f\in L^\infty(X,\mu)$ and $b\in B$, let $f[b]\subset\mathbb{C}$ denote the essential range of $f$ on $b$, and let $C_f=\{f[b]:b\in U\}$. Note that each element of $C_f$ is compact and nonempty. Also, $f[b\cap c]\subseteq f[b]\cap f[c]$, so $C_f$ has the finite intersection property. Thus $\bigcap C_f$ is nonempty. If $x\in \bigcap C_f$, then for any neighborhood $V$ of $x$ and any $b\in U$, $f^{-1}(V)\cap b$ is non-null. Since $U$ is an ultrafilter on $B$, this means $f^{-1}(V)\in U$. Now if we had two different points $x,y\in C_f$, they would have disjoint neighbooods $V$ and $W$, and then $f^{-1}(V)$ and $f^{-1}(W)$ would be disjoint elements of $U$. This is impossible.
Thus, we have shown that $C_f$ has exactly one point for each $f\in L^\infty(X,\mu)$. Define $\alpha(f)$ to be the unique element of $C_f$, which can also be described as the unique point $x$ such that given any neighborhood $V$ of $x$, for all sufficiently small $b\in U$, $f|_b$ takes values in $V$ almost everywhere. This description makes it easy to verify that $\alpha$ is a $*$-homomorphism, and that $\alpha(1_b)=h(b)$ for each $b\in B$. Thus $\alpha\in Y$ and $h=F(\alpha)$, so $h$ is in the image of $F$, as desired.
(Alternatively, to show $F$ is surjective, by Stone duality it suffices to show that the image of $F$ separates elements of $B$, since closed subspaces of the Stone space $S$ correspond to quotients of the algebra $B$. But by Gelfand duality, elements of $Y$ separate elements of $L^\infty(X,\mu)$, and so we're done since distinct elements of $B$ have distinct characteristic functions in $L^\infty(X,\mu)$.)
Best Answer
Let me try another perhaps more pedestrian approach.
So we're given that $M=\ell ^\infty (M_i)$ (I guess I prefer this notation over ${\oplus}_i^\infty M_i$), where the $M_i$ are von Neumann algebras. Denoting by $M_*$ the space of normal linear functionals, we thus need to prove that $$ M_*=\ell ^1({M_i}_*). $$
Given $\omega \in M_*$, denote by $\omega _i$ the restriction of $\omega $ to $M_i$, so that $\omega _i\in {M_i}_*$, and let us prove that $$ \|\omega \| = \sum_i\Vert \omega _i\Vert. \tag 1 $$
Fixing $\rho $ in the interval $(0, 1)$, and for each $i$, choose some $m_i$ in $M_i$ such that $\Vert m_i\Vert =1$, and $\omega _i(m_i)\geq \rho \Vert \omega _i\Vert $.
If $F$ is a finite subset of the index set $I$, and $m_F\in M$ is defined to have coordinates $m_i$, for $i$ in $F$, and zero elsewhere, then $\Vert m_F\Vert =1$, and $$ \Vert \omega \Vert \geq |\omega (m_F)| = \sum_{i\in F}\omega _i(m_i) \geq \sum_{i\in F} \rho \Vert \omega _i\Vert . $$ Since both $F$ and $\rho $ are arbitrary, it follows that $\Vert \omega \Vert \geq \sum_{i\in I} \Vert \omega _i\Vert $.
In order to prove the reverse inequality, let $N$ be the subalgebra of $M$ formed by the elements with finitely many nonzero coordinates. Then, for every $m$ in the unit ball of $N$, we have that $$ |\omega (m)| = \big |\sum_{i\in I}\omega _i(m_i)\big | \leq \sum_{i\in I}|\omega _i(m_i) | \leq $$$$ \leq \sup_{i\in I}\|m_i\|\sum_{i\in I}\|\omega _i\| \leq \sum_{i\in I}\|\omega _i\|. \tag 2 $$
Using von Neumann's double commutant Theorem, it is easy to see that $N$ is weakly dense in $M$. Furthermore, by Kaplansky's density theorem, we have that the unit ball of $N$ is dense in the unit ball of $M$ relative to $\sigma (M,M_*)$, aka the weak$^*$ topology. Therefore the inequality in (2) is preserved if, instead, $m$ is taken in the unit ball of $M$. This concludes the proof of claim (1)
Therefore this defines an isometric linear map $$ \omega \in M_*\mapsto (\omega _i)_i\in \ell ^1({M_i}_*), \tag 3 $$ and we'll be done once we prove it to be onto.
To see that this is indeed the case, pick $(\omega _i)_i\in \ell ^1({M_i}_*)$, and define for each subset $J\subseteq I$, the linear functional $\omega _J$ on $M$ by $$ \omega _J(m) =\sum_{i\in F} \omega _i(m_i),\quad\forall m\in M. $$
For $J$ finite it is clear that $\omega _J$ is normal, and hence so is $\omega :=\omega _I$, since the latter is the norm limit of the $\omega _F$, for $F$ finite.
Finally notice that the map in (3) sends $\omega $ to $(\omega _i)_i$, concluding the proof.