[proofplan] The bounded direct sum has coordinatewise multiplication, so centrality in $M$ should be equivalent to centrality in every coordinate. To prove the nontrivial direction, we test a central element against elements supported at a single index, which are allowed because the direct sum is bounded rather than algebraic or $c_0$. The converse follows by coordinatewise multiplication. In the factor case, each coordinate center is $\mathbb C1_i$, and the coordinate scalars give the explicit isomorphism with $\ell^\infty(I)$. [/proofplan]

[step:Translate centrality in the direct sum into coordinatewise commutation]Let $x=(x_i)_{i\in I}\in Z(M)$. Fix an index $k\in I$ and an element $a\in M_k$. Define $y^{(k,a)}=(y_i)_{i\in I}\in M$ by \begin{align*} y_i = \begin{cases} a, & i=k, \end{align*} \begin{align*} 0, & i\ne k. \end{cases} \end{align*} This family belongs to $M$ because \begin{align*} \sup_{i\in I}\|y_i\|_{\mathrm{op}}=\|a\|_{\mathrm{op}}<\infty. \end{align*} Since $x\in Z(M)$, we have $xy^{(k,a)}=y^{(k,a)}x$. Taking the $k$-th coordinate and using coordinatewise multiplication gives \begin{align*} x_k a=a x_k. \end{align*} Because $a\in M_k$ was arbitrary, $x_k\in Z(M_k)$. Since $k\in I$ was arbitrary, $x_i\in Z(M_i)$ for every $i\in I$. The original boundedness of $x\in M$ gives \begin{align*} \sup_{i\in I}\|x_i\|_{\mathrm{op}}<\infty, \end{align*} so $x\in\bigoplus_{i\in I}Z(M_i)$.[/step]

[guided]We start with an element $x=(x_i)_{i\in I}\in Z(M)$ and must prove that every coordinate $x_i$ is central in its own algebra $M_i$. Fix one coordinate, denoted by $k\in I$, and fix an arbitrary element $a\in M_k$. The goal is to prove that $x_k a=a x_k$. To isolate the $k$-th coordinate, define a family $y^{(k,a)}=(y_i)_{i\in I}$ by \begin{align*} y_i = \begin{cases} a, & i=k, \end{align*} \begin{align*} 0, & i\ne k. \end{cases} \end{align*} This is an element of the bounded direct sum $M$, not merely a formal family, because its coordinate norms are bounded: \begin{align*} \sup_{i\in I}\|y_i\|_{\mathrm{op}}=\|a\|_{\mathrm{op}}<\infty. \end{align*} This is exactly where the phrase “bounded direct sum” matters: elements supported at one coordinate are permitted. Since $x$ lies in the center $Z(M)$, it commutes with every element of $M$, and in particular with $y^{(k,a)}$. Thus \begin{align*} xy^{(k,a)}=y^{(k,a)}x. \end{align*} Multiplication in $M$ is coordinatewise, so the $k$-th coordinate of the left-hand side is $x_k a$, while the $k$-th coordinate of the right-hand side is $a x_k$. Therefore \begin{align*} x_k a=a x_k. \end{align*} The element $a\in M_k$ was arbitrary, so $x_k$ commutes with every element of $M_k$. Hence $x_k\in Z(M_k)$. Since the coordinate $k\in I$ was arbitrary, $x_i\in Z(M_i)$ for every $i\in I$. Finally, because $x$ was already an element of $M$, its coordinates are uniformly bounded: \begin{align*} \sup_{i\in I}\|x_i\|_{\mathrm{op}}<\infty. \end{align*} Thus $x$ is a bounded family of elements with $x_i\in Z(M_i)$ for every $i$, which is precisely the statement that \begin{align*} x\in\bigoplus_{i\in I}Z(M_i). \end{align*}[/guided]

[step:Show that bounded families of coordinatewise central elements are central] Let $x=(x_i)_{i\in I}\in\bigoplus_{i\in I}Z(M_i)$. Then $x\in M$, and for every $y=(y_i)_{i\in I}\in M$ we have $y_i\in M_i$ for each $i\in I$. Since $x_i\in Z(M_i)$, coordinatewise multiplication gives \begin{align*} (xy)_i=x_i y_i=y_i x_i=(yx)_i \end{align*} for every $i\in I$. Hence $xy=yx$ for every $y\in M$, so $x\in Z(M)$. Combining this with the previous step gives \begin{align*} Z(M)=\bigoplus_{i\in I}Z(M_i). \end{align*} [/step]

[step:Identify the center with $\ell^\infty(I)$ when all summands are factors] Assume now that every $M_i$ is a factor. By the definition of a factor, \begin{align*} Z(M_i)=\mathbb C1_i \end{align*} for every $i\in I$. Define \begin{align*} \Phi:\ell^\infty(I)&\to Z(M) \end{align*} \begin{align*} (\lambda_i)_{i\in I}&\mapsto (\lambda_i1_i)_{i\in I}. \end{align*} This map is well-defined because \begin{align*} \sup_{i\in I}\|\lambda_i1_i\|_{\mathrm{op}}=\sup_{i\in I}|\lambda_i|<\infty. \end{align*} It is linear, multiplicative, preserves the involution, and sends the constant function $1$ to the unit $(1_i)_{i\in I}$ of $M$, all by coordinatewise operations. It is injective because $\lambda_i1_i=0$ implies $\lambda_i=0$ for every $i\in I$, and it is surjective because every element of $Z(M)$ is a bounded family $(\lambda_i1_i)_{i\in I}$ by the already proved equality \begin{align*} Z(M)=\bigoplus_{i\in I}Z(M_i). \end{align*} Therefore $\Phi$ is a $*$-isomorphism of von Neumann algebras, and hence \begin{align*} Z(M)\cong \ell^\infty(I). \end{align*} This proves the theorem. [/step]