Suppose $S$ is non-abelian simple. Using https://groupprops.subwiki.org/wiki/Normal_subdirect_product_of_perfect_groups_equals_direct_product we know the normal subgroups of $S^n$ are of the form $S^I$ for some subset $I\subset\{1,\dots,n\}$ of indices. So $\varphi$ being irreducible is equivalent to $\varphi(S^I)\not\subset S^I$.

Next, note the following fact that Qiaochu mentions in the comments:

Lemma The automorphism group of $S^n$ is $\mathrm{Aut}(S)\wr \Sigma_n$.

Proof Let $\varphi\colon S^n\to S^n$ be an automorphism. For each $i\in\{1,\dots,n\}$, the $i$-th summand $S_i\subset S^n$ is a normal subgroup, hence $\varphi(S_i)\subset S^n$ is a normal subgroup. But by the above characterization of normal subgroups of $S^n$ we know $\varphi(S_i)=S_{\sigma(i)}$ for some index $\sigma(i)\in\{1,\dots,n\}$. Since $\varphi$ is an automorphism we know that $\sigma$ is a bijection, hence any automorphism $\phi$ breaks up into a permutation of the factors and an automorphism of $S$ on each factor.

Now, let $\varphi\in\mathrm{Aut}(S^n)$ be an automorphism which induces a permutation $\sigma\in \Sigma_n$ on the direct factors. Then the condition that $\varphi(S^I)\not\subset S^I$ is equivalent to $\sigma(I)\not\subset I$ for any non-trivial subset $I$ of $\{1,\dots,n\}$. This is equivalent to $\sigma$ being a $n$-cycle. Thus, we conclude:

Proposition An automorphism $\varphi$ of $S^n$ is irreducible if and only if the permutation $\sigma\in\Sigma_n$ induced on the direct factors is a $n$-cycle.

Suppose $S$ is non-abelian simple. Using https://groupprops.subwiki.org/wiki/Normal_subdirect_product_of_perfect_groups_equals_direct_product we know the normal subgroups of $S^n$ are of the form $S^I$ for some subset $I\subset\{1,\dots,n\}$ of indices. So $\varphi$ being irreducible is equivalent to $\varphi(S^I)\not\subset S^I$.

Next, note the following fact that Qiaochu mentions in the comments:

Lemma The automorphism group of $S^n$ is $\mathrm{Aut}(S)\wr \Sigma_n$.

Proof Let $\varphi\colon S^n\to S^n$ be an automorphism. For each $i\in\{1,\dots,n\}$, the $i$-th summand $S_i\subset S^n$ is a normal subgroup, hence $\varphi(S_i)\subset S^n$ is a normal subgroup. But by the above characterization of normal subgroups of $S^n$ we know $\varphi(S_i)=S_{\sigma(i)}$ for some index $\sigma(i)\in\{1,\dots,n\}$. Thus $\varphi=\psi\circ \sigma$ where $\sigma$ permutes the factors and $\psi$ is an automorphism of $S^n$ sending $S_i$ to $S_i$. But then $\psi=\psi_1\times\cdots\times\psi_n$ where $\psi_i\in\mathrm{Aut}(S)$.

Now, let $\varphi\in\mathrm{Aut}(S^n)$ be an automorphism which induces a permutation $\sigma\in \Sigma_n$ on the direct factors. Then the condition that $\varphi(S^I)\not\subset S^I$ is equivalent to $\sigma(I)\not\subset I$ for any non-trivial subset $I$ of $\{1,\dots,n\}$. This is equivalent to $\sigma$ being a $n$-cycle. Thus, we conclude:

Proposition An automorphism $\varphi$ of $S^n$ is irreducible if and only if the permutation $\sigma\in\Sigma_n$ induced on the direct factors is a $n$-cycle.