When the group order is invertible the Reynolds operator retracts a ring onto its invariants

The element $u=N\cdot 1_S$ is fixed by the action, and so is its inverse. Each $h\in G$ acts as a ring homomorphism, so it is additive and sends $1_S$ to $1_S$; applying it to the $N$-fold sum $1_S+\cdots +1_S$ gives $h\cdot u=u$. Applying $h$ to $u{u}^{-1}=1_S$ gives $u(h\cdot u^{-1})=1_S$; left-multiplying by $u^{-1}$ and using associativity, $u^{-1}u=1_S$, and $1_{S}x=x$ yields $h\cdot u^{-1}=u^{-1}$. The formula $\rho (s)=u^{-1}{\sum }_{g\in G}g\cdot s$ therefore defines a function $S\to S$, the sum being over the finite set $G$.

$\rho$ takes values in $S^G$. For $h\in G$, additivity of the action of $h$ and step 1.1 give $h\cdot \rho (s)=u^{-1}{\sum }_{g\in G}h\cdot (g\cdot s)=u^{-1}{\sum }_{g\in G}(hg)\cdot s$; and $g\mapsto hg$ is a bijection of $G$ onto itself, with inverse $g\mapsto h^{-1}g$, so it merely reindexes the sum and $h\cdot \rho (s)=\rho (s)$.

$\rho$ is $S^G$-linear. Additivity is additivity of each $g$ together with associativity and commutativity of addition. For $a\in S^G$ and $s\in S$, each $g$ is multiplicative and fixes $a$, so $g\cdot (as)=(g\cdot a)(g\cdot s)=a(g\cdot s)$; summing and using distributivity gives $\rho (as)=u^{-1}a{\sum }_g(g\cdot s)=a\rho (s)$, where $a$ and $u^{-1}$ commute because $S$ is commutative.

$\rho$ fixes $S^G$ pointwise. For $a\in S^G$ every term of the sum is $a$, so ${\sum }_{g\in G}g\cdot a$ is the $N$-fold sum of $a$, which by distributivity is $ua$; hence $\rho (a)=u^{-1}(ua)=a$.

So $S^G$ is a subring of $S$ and $\rho \text{\hspace{0.17em}⁣}:S\to S^G$ is $S^G$-linear with $\rho (a)=a$ for every $a\in S^G$: it is a retraction of the inclusion as a map of $S^G$-modules. If $S$ is Noetherian, the retraction lemma applies with $R^{\mathrm{\prime }}=S$ and $R=S^G$ and gives that $S^G$ is Noetherian.

The comparison with Noether's theorem, and the caveat. Noether's theorem needs a Noetherian subring $A\subseteq C$, the finite-type hypothesis over $A$, and an action by $A$-algebra automorphisms, and it concludes that $C^G$ is of finite type over $A$. The argument here uses none of the finite-type or fixed-base-ring hypotheses and concludes only that $S^G$ is Noetherian, at the cost of the invertibility of $u$. That cost is real: in $S={\mathbb{F}}_2[x]$, the substitution $x\mapsto x+1$ defines an automorphism of order $2$, but for the resulting action of the order-two group one has $u=2\cdot 1_S=0$, so $\rho$ is not defined.