[proofplan] We expand the cumulant of the sum by multilinearity of the free cumulant functional in its $n$ arguments. The expansion is indexed by subsets of $\{1,\dots,n\}$, according to which slots contain $a$ and which slots contain $b$. The two terms with only one variable are exactly the two marginal cumulants, and every other term is a mixed cumulant involving elements from two free unital subalgebras. Those mixed terms vanish by the standard mixed-cumulant characterization of freeness. [/proofplan]

[step:Expand the cumulant by multilinearity over all choices of slots]Fix $n \in \mathbb N$ and define the finite index set $I_n := \{1,\dots,n\}$. For each subset $S \subset I_n$ and each $i \in I_n$, define $x_{i,S} \in \mathcal A$ by declaring $x_{i,S}=a$ when $i \in S$ and $x_{i,S}=b$ when $i \notin S$. Since $\kappa_n: \mathcal A^n \to \mathbb C$ is multilinear in its $n$ arguments, expanding each copy of $a+b$ gives \begin{align*} \kappa_n(a+b,\dots,a+b)=\sum_{S \subset I_n}\kappa_n(x_{1,S},\dots,x_{n,S}). \end{align*}[/step]

[guided]Fix $n \in \mathbb N$. We need a compact way to record every term that appears when each of the $n$ entries $a+b$ is expanded. Define the finite index set $I_n := \{1,\dots,n\}$. For a subset $S \subset I_n$, the convention is that the slots in $S$ receive $a$, and the slots outside $S$ receive $b$. Formally, for each $i \in I_n$, define $x_{i,S} \in \mathcal A$ by declaring $x_{i,S}=a$ when $i \in S$ and $x_{i,S}=b$ when $i \notin S$. This definition accounts for every possible choice of either $a$ or $b$ in each argument of $\kappa_n$. The free cumulant functional $\kappa_n: \mathcal A^n \to \mathbb C$ is multilinear, meaning that it is linear separately in each one of its $n$ entries. Applying that separate linearity successively to the first entry, then to the second entry, and continuing through the $n$-th entry gives the full finite expansion \begin{align*} \kappa_n(a+b,\dots,a+b)=\sum_{S \subset I_n}\kappa_n(x_{1,S},\dots,x_{n,S}). \end{align*} The sum is finite because $I_n$ has exactly $n$ elements, so there are exactly $2^n$ subsets $S \subset I_n$.[/guided]

[step:Separate the unmixed cumulants from the mixed cumulants]The subset $S=I_n$ gives $x_{i,I_n}=a$ for every $i \in I_n$, hence \begin{align*} \kappa_n(x_{1,I_n},\dots,x_{n,I_n})=\kappa_n(a,\dots,a). \end{align*} The subset $S=\varnothing$ gives $x_{i,\varnothing}=b$ for every $i \in I_n$, hence \begin{align*} \kappa_n(x_{1,\varnothing},\dots,x_{n,\varnothing})=\kappa_n(b,\dots,b). \end{align*} If $S$ is neither $\varnothing$ nor $I_n$, then there exist indices $i,j \in I_n$ such that $i \in S$ and $j \notin S$. Therefore the tuple $(x_{1,S},\dots,x_{n,S})$ contains at least one entry equal to $a \in \mathcal B$ and at least one entry equal to $b \in \mathcal C$.[/step]

[guided]The expansion is indexed by all subsets $S \subset I_n$, so we now identify which subsets give the two terms that should remain. If $S=I_n$, then every slot is marked as an $a$-slot. Thus $x_{i,I_n}=a$ for every $i \in I_n$, and therefore \begin{align*} \kappa_n(x_{1,I_n},\dots,x_{n,I_n})=\kappa_n(a,\dots,a). \end{align*} If $S=\varnothing$, then no slot is marked as an $a$-slot, so every slot receives $b$. Thus $x_{i,\varnothing}=b$ for every $i \in I_n$, and hence \begin{align*} \kappa_n(x_{1,\varnothing},\dots,x_{n,\varnothing})=\kappa_n(b,\dots,b). \end{align*} These are precisely the two unmixed cumulants appearing in the desired identity. It remains to classify every other subset. Let $S \subset I_n$ satisfy $S \neq \varnothing$ and $S \neq I_n$. Since $S$ is nonempty, there is an index $i \in I_n$ with $i \in S$, so $x_{i,S}=a \in \mathcal B$. Since $S$ is not all of $I_n$, there is an index $j \in I_n$ with $j \notin S$, so $x_{j,S}=b \in \mathcal C$. Hence the tuple $(x_{1,S},\dots,x_{n,S})$ contains entries from both free unital subalgebras $\mathcal B$ and $\mathcal C$. These are exactly the mixed terms that freeness will eliminate in the next step.[/guided]

[step:Use freeness to eliminate every mixed cumulant]Let $S \subset I_n$ satisfy $S \neq \varnothing$ and $S \neq I_n$. Then $n \geq 2$, because $I_n$ contains at least one index in $S$ and at least one index outside $S$. By the preceding step, the tuple $(x_{1,S},\dots,x_{n,S})$ contains entries labelled by both $\mathcal B$ and $\mathcal C$: slots in $S$ are assigned $a \in \mathcal B$, and slots in $I_n \setminus S$ are assigned $b \in \mathcal C$. We use the mixed free cumulant characterization of freeness: if $\mathcal B$ and $\mathcal C$ are free unital subalgebras and $m \geq 2$, then every scalar free cumulant $\kappa_m(y_1,\dots,y_m)$ vanishes whenever each $y_i$ is chosen from one of the free subalgebras according to a nonconstant subalgebra-label map. Its hypotheses hold here with $m=n$, because $\mathcal B$ and $\mathcal C$ are free by assumption and the label map determined by membership in $S$ is nonconstant. Thus \begin{align*} \kappa_n(x_{1,S},\dots,x_{n,S})=0 \end{align*} for every proper nonempty subset $S \subset I_n$. Substituting this vanishing into the multilinear expansion leaves only the two unmixed terms: \begin{align*} \kappa_n(a+b,\dots,a+b)=\kappa_n(a,\dots,a)+\kappa_n(b,\dots,b). \end{align*} This is the desired identity for the fixed $n \in \mathbb N$, and since $n$ was arbitrary, the result holds for every $n \geq 1$.[/step]

[guided]Let $S \subset I_n$ be a proper nonempty subset. The words "proper" and "nonempty" matter: nonempty gives an index $i \in S$, while proper gives an index $j \in I_n \setminus S$. Therefore $I_n$ has at least two elements, so $n \geq 2$. By the definition of $x_{i,S}$, the tuple $(x_{1,S},\dots,x_{n,S})$ contains at least one entry labelled by $\mathcal B$, namely $a \in \mathcal B$, and at least one entry labelled by $\mathcal C$, namely $b \in \mathcal C$. We now invoke the precise external result needed here, the mixed free cumulant characterization of freeness. It says that if $\mathcal B$ and $\mathcal C$ are free unital subalgebras and $m \geq 2$, then every scalar free cumulant $\kappa_m(y_1,\dots,y_m)$ vanishes whenever each $y_i$ is chosen from one of the free subalgebras and the associated subalgebra-label map is nonconstant. We verify the hypotheses in the present situation. The order condition holds with $m=n$ because a proper nonempty $S$ can occur only when $n \geq 2$. The freeness hypothesis holds because $\mathcal B$ and $\mathcal C$ are assumed free. Finally, each entry $x_{i,S}$ lies in either $\mathcal B$ or $\mathcal C$, and the label map is nonconstant because the slots in $S$ are assigned to $\mathcal B$ while the slots in $I_n \setminus S$ are assigned to $\mathcal C$. Hence the criterion gives \begin{align*} \kappa_n(x_{1,S},\dots,x_{n,S})=0. \end{align*} Since this argument applies to every proper nonempty subset $S \subset I_n$, all mixed terms in the multilinear expansion vanish. The expansion from the first step was \begin{align*} \kappa_n(a+b,\dots,a+b)=\sum_{S \subset I_n}\kappa_n(x_{1,S},\dots,x_{n,S}). \end{align*} The subset $S=I_n$ contributes $\kappa_n(a,\dots,a)$, the subset $S=\varnothing$ contributes $\kappa_n(b,\dots,b)$, and every other subset contributes zero. Therefore \begin{align*} \kappa_n(a+b,\dots,a+b)=\kappa_n(a,\dots,a)+\kappa_n(b,\dots,b). \end{align*} This proves the identity for the fixed $n \in \mathbb N$. Since the choice of $n$ was arbitrary, the identity holds for every $n \geq 1$.[/guided]