All the rings will be commutative with 1.
Definition. Let $S/R$ be a ring extension (that is, $R$ is a subring of $S$), and let $G$ be a (finite) subgroup of $\operatorname{Aut}_{R-\text{alg}}(S)$. We say that $S/R$ is a Galois extension of group $G$ if $R=S^G$ and one of the following equivalent condition holds:
(i) The map $x\otimes y\in S\otimes_R S\mapsto (x\sigma(y))_{\sigma\in G}\in S^{\vert G\vert}$ is an isomorphism of $R$-algebras
(ii) There exists $x_1,y_1,...,x_n,y_n\in S$ such that $\sum_{i=1}^n x_i\sigma(y_i)=\delta_{\sigma,Id_S}$ for all $\sigma\in G$
(iii) For each maximal ideal $\mathfrak{M}$ of $S$ and each $\sigma\in G\setminus\{Id_S\}, $there exists $x\in S$ such that $\sigma(x)-x\notin \mathfrak{M}$.
There are also equivalent definitions involving trivial crossed product algebras, but i do not want to come into that.
Context. The following result is well-known: if ${\rm Pic}(R)=0$, then $H^1(G,S^\times)=1$ (see for example, Ford, separable algebras, Corollary 13.3.2)
In particular, for cyclic groups, the well-known description of the cohomology of cyclic groups yields :
Prop. Assume that $S/R$ is Galois, with cyclic Galois group. Let $\sigma$ be a generator of the Galois group. Assume that ${\rm Pic}(R)=0$
Then, for all $x\in S^\times$, we have $N(x)=1$ if and only if there exists $y\in S^\times$ such that $x=\dfrac{y}{\sigma(y)}$.
Here, $N(x)=x\sigma(x)\cdots \sigma^{n-1}(x)$, where $n=o(\sigma)$.
The problem. For various reasons, I would like a direct proof of the previous result when $R$ is semi-local.
I tried to adapt the standard proof , known for cyclic field extensions, so the problem boils down to the following question.
Question. Assuming that $R$ is a semi-local ring, and given $x\in S^\times$ such that $N(x)=1$, does there exists $z\in S$ such that the quantity $y=z+x\sigma(z)+(x\sigma(x))\sigma^2(z)+\cdots+(x\sigma(x)\cdots\sigma^{n-2}(x))\sigma^{n-1}(z)$ is a unit of $S$ ?
If it makes things simpler, I will be satisfied to get an elementary argument in the following special case: $S=L\otimes_F R$, where $L$ and $F$ are finite fields and $R$ is a finite dimensional commutative $F$-algebra (hence $R$ and $S$ are also finite)
Of course I will be happy with any other elementary method to get the desired result.