Let $K$ be a field and let $E/K$ be a Galois extension of degree 6 with $\text{Gal}(E/K) = S_3$, the symmetric group on 3 letters. Pick two different transpositions $s_1, s_2$ in $S_3$ (hence $s_1s_2$ is of order $3$). Let $x\in E$ be such that $s_1 (x) = s_2(x) = x^{-1}$ (in particular, such an $x$ belongs to the quadratic sub-extension $F$ of E/K, since it is ($s_1s_2$)-invariant. Another way to put it is that $ x=\frac{\bar{a}}{a}$ for some $a\in F^{\ast}$, where $\bar{.}$ denotes the conjugation in $\text{Gal}(F/K)$). Finally assume that there does not exist $y\in E$ such that $y.(s_1s_2)(y).(s_1s_2)^2(y)=x$.

QUESTION: How to prove that there exists a degree $3$ extension $L$ of $E$ which is Galois over $K$ with Galois group $\mathbf{Z}/3\mathbf{Z}\times S_3$ and such that there exists $y\in L$ satisfying $y.(s_1s_2)(y).(s_1s_2)^2(y)=x$?

(In the question, I use the given splitting $\text{Gal}(E/K)\to \text{Gal}(L/K)$ to see $s_i$ acting on $L$).

Background: This question arose whilst trying to check in a cohomological manner that division algebras of degree $3$ are cyclic, a theorem of Wedderburn from 1921. Cyclicity of degree $3$ division algebras should imply the existence of $L$ as in the question (that would be one way to prove it, but I'm looking for other ways). I think I can handle the question when $E=K[\sqrt[3]{a}]$ (where $a\in K$ and $K$ does not have a primitive $3$rd root of unity), but I'm lost otherwise.

EDIT: I can't even tackle the cases of a number field or of a local field, so I added the tags "number theory" and "local fields" to get some help in these cases.