Question: Is $S_{n}(X) \oplus S_{n}(Y) \cong S_{n}(X \amalg Y)$?
Proof attempt: Here's my argument: Suppose that $X$ and $Y$ are disjoint topological spaces.
Note that if $\sigma : \Delta^{n} \rightarrow X \amalg Y$ is an $n$-simplex, then $\sigma(\Delta^{n})$ is connected since $\Delta^{n}$ is connected. Thus either $\sigma(\Delta^{n}) \subseteq X$ or $\sigma(\Delta^{n}) \subseteq Y$.
This gives us an injective map defined by $f_{X} : \sigma \in \text{Sin}_{n}(X) \mapsto \iota_{X} \circ \sigma \in \text{Sin}_{n}(X \amalg Y)$, where $\iota_{X}: X \rightarrow X \amalg Y$ is the inclusion map. Likewise, an injective map $f_{Y}$ exists.
Observe that $f_{X}$ extends to a homomorphism. In particular, if $\sigma_{1}$ and $\sigma_{2}$ are $n$-simplices in $X$ and $n_{1},n_{2} \in \mathbb{Z}$, we define $f_{X} : S_{n}(X) \rightarrow S_{n}(X \amalg Y)$ by $$f_{X}(n_{1} \sigma_{1} + n_{2} \sigma_{2}) := n_{1}(\iota_{X} \circ \sigma_{1}) + n_{2}(\iota_{X} \circ \sigma_{2}).$$ Likewise for $f_{Y}$.
Taking $S_{n}(X) \oplus S_{n}(Y)$, we get a homomorphism $f_{X} \oplus f_{Y} : S_{n}(X) \oplus S_{n}(Y) \rightarrow S_{n}(X \amalg Y)$ defined by $f_{X} \oplus f_{Y}(\sigma_{X}, \sigma_{Y}) := f_{X}(\sigma_{X}) + f_{Y}(\sigma_{Y})$, whose inverse is given by the map $\sigma'_{X} + \sigma'_{Y} \mapsto (\sigma'_{X}, \sigma'_{Y})$, where $\sigma'_{X}$ is an $n$-simplex in the subspace $X$ of $X \amalg Y$, and $\sigma'_{Y}$ is an $n$-simplex in a subspace $Y$ of $X \amalg Y$.
Comments: Notice here that my argument relies on $\Delta^{n}$ being connected. Although I haven't proved it, I thought it is obvious. I appreciate it if someone can correct me. And I would appreciate it if someone can point out the errors that I might have missed in the functions that I defined.
Your proof is correct. Only a little bit of nitpicking: You say "Suppose that $X$ and $Y$ are disjoint topological spaces." This is not what is needed, the disjoint union of $X \amalg Y$ is defined for any two spaces $X,Y$. Technically you can understand it as the set $X \times \{1\} \cup Y \times \{2\}$ with open subsets of the form $U \times \{1\}$ and $V \times \{2\}$, where $U \subset X$ and $V \subset Y$ are open. The homeomorphic copies $X' = X \times \{1\}$ of $X$ and $Y' = Y \times \{2\}$ of $Y$ are disjoint open subsets of $X \amalg Y$, and this is what you need to conclude that either $\sigma(\Delta^{n}) \subseteq X'$ or $\sigma(\Delta^{n}) \subseteq Y'$. Note that if $X \cap Y = \emptyset$, you can in general not take $X \amalg Y = X \cup Y$. As an example take $X= [0,1]$ and $Y = (1,2]$.