Let $E=\mathbb{F}_p(\!(u)\!)$, the Laurent series field over $\mathbb{F}_p$. Let $K/E$ be a finite normal separable extension. Consider the field $L=K(x \mid x^p-x-a=0 \text{ for some } a \in K)$. Let $\hat{L}$ be the $u$-adic completion of $L$. We write $G_E=\mathrm{Gal}(E^s/E)$ for the absolute Galois group of $E$. By continuity $G_E$ acts on $\hat{L}$. What is $(\hat{L})^{G_E}$?

In particular, is $(\hat{L})^{G_E}= E$? One has $\hat{L} \subseteq \widehat{E^s}$ and following the answer to one of my last questions one also has $(\widehat{E^s})^{G_E}=\widehat{E^{\mathrm{perf}}}$. Therefore, we are left with the question if $\hat{L} \cap \widehat{E^{\mathrm{perf}}}=E$?

Edit: and what if we consider for some $m \in \mathbb{Z}$ the field $\hat{L}_m$ where $L_m= K(x \mid x^p-x-a=0 \text{ for some } a \in K \text{ with } \nu(a) \geq m)$? What in the special case where $m=0$?