Aluffi II.6.11 suggests proving the above.
Here's the sketch of my proof by contradiction.
Assume that $S_3$ is a coproduct of a family $\mathcal{C}$ of cyclic groups $C^i$. Using the universal property for coproducts (and considering morphisms $\sigma_i : S_3 \rightarrow C^i$ such that $\sigma_i \iota_i = 1$, where $\iota_i$ is the injection function) shows that each $C^i$ cannot be larger (as a set) than $S_3$.
Next, since $S_3$ has three elements of order 2 and two elements of order 3, it can be shown that any cyclic group that maps its generator onto one of those elements shall have order 2 or 3 respectively.
So we're down to a corpoduct of a certain number of $C_2$ and $C_3$.
Now, by considering both elements of order 3 in $S_3$, it can be shown that having two $C_3 \in \mathcal{C}$ with different injection functions leads to contradiction, so we have at most one $C_3$. We also have to have at least one $C_2$ mapping onto some element of order 2, otherwise there's a certain freedom in defining the behavior of some morphisms from $S_3$. Similarly, we also have to have at least one $C_3$.
Now, considering the group $C_2 \times C_3$ along with a pair of morphisms $\varphi_2 : C_2 \rightarrow C_2 \times C_3, \varphi_2([n]_2) = ([n]_2, [0]_3)$ and $\varphi_3 : C_3 \rightarrow C_2 \times C_3, \varphi_3([n]_3) = ([0]_2, [n]_3)$ it can be shown that there is no valid homomorphism $\sigma : S_3 \rightarrow C_2 \times C_3$ satisfying the corresponding universal property for coproducts, hence the contradiction with the original assumption.
Overall, this looks quite clunky. Does it look reasonable though? Is there a better way to prove the claim (perhaps limited to the little amount of algebra and category theory that might be expected by this point)?
Best Answer
Here is a maybe simpler proof, but which involves more concepts that may be unknown.
$\newcommand{\Z}{\mathbb{Z}}$Suppose that a group $G$ is the coproduct of cyclic groups: $$G = \Z/n_1\Z * \dots * \Z/n_k\Z.$$ Then the abelianization of $G$, i.e. its quotient by its commutator subgroup, is isomorphic to the direct sum $G_{\mathrm{ab}} \cong \Z/n_1\Z \oplus \dots \oplus \Z/n_k\Z$. So if we know the abelianization, we know what cyclic groups must be involved.
Now it's well-known that the commutator subgroup of $S_3$ is the alternating group $A_3$, and the quotient $(S_3)_{\mathrm{ab}}$ is the cyclic group $\Z/2\Z$. Therefore, if $S_3$ were a coproduct of cyclic groups, then it would be isomorphic to $\Z/2\Z$. This is obviously false because $S_3$ has $6$ elements whereas $\Z/2\Z$ only has two. Therefore $S_3$ cannot be isomorphic to a coproduct of cyclic groups.