Just to expand a bit on Dalawat's answer, let me make a couple of observations. First, there is a very explicit description of $U(1)$, namely
$$
U(1)=\{x\in E\text{ such that }x=\frac{y}{\sigma y}\text{ for some }y\in E\}
$$
coming from Hilbert 90 which says $H^1(E/F,E^\times)=1$ coupled with the information that, since the Galois group here is cyclic$^\dagger$ there is an isomorphism
$$
H^1(E/F,E^\times)\cong \hat{H}^{-1}(E/F,E^\times)\stackrel{\text{def}}{=}U(1)/\{y/\sigma y: y\in E\}.
$$
This already tells you something but there is more to say: namely, we could try to describe $U(1)$ in terms of units only, not using non-invertible elements in $E^\times$. Then a classical result (see for instance Serre's paper in Cassels-Fröhlich) tells you that the cohomology of units in *unramified* extensions is trivial, so the same argument above becomes
$$
U(1)=\{x\in U_E\text{ such that }x=\frac{y}{\sigma y}\text{ for some }y\in U_E\}\quad\text{ if }E/F\text{ unramified}.
$$
If $E/F$ is ramified, there are two options: either $p=2$ or $p>2$. In the second case, the restriction-corestriction trick kills a lot of the cohomology of the units: writing $U_E=U'\times\mu_E$ with $U'$ being the principal units which are $1$ modulo the maximal ideal, we know $U'$ is a pro-$p$ group and hence has trivial cohomology. We find
$$
H^1(E/F,\mu_E)\cong\hat{H}^{-1}(E/F,\mu_E)\stackrel{\text{res-cores}}{\cong}\hat{H}^{-1}(E/F,U_E)\stackrel{\text{def}}{=}U(1)/\{y/\sigma y: y\in U_E\}.
$$
and you see that "up to the finite group $H^1$ of roots of unity" the answer is still the same, namely that the kernel of norm is the same as elements $\sigma y/y$.

The above fails is $E/F$ is ramified and $p=2$ (the wildly ramified case Dalawat's is referring to) but in Serre's paper quoted above you will find a Lemma (I am sorry for being unable to give proper references, I do not have my Cassels-Fröhlich at hand) telling you that my argument still holds by replacing the whole $U_E$ with a smaller, but still finite-index, subgroup and hence the vague sentence that "up to a finite group the kernel of norm is the same as elements $\sigma y/y$" still holds. The exact determination of this finite group in general depends upon the extension $E/F$.

$^\dagger$: Purists would say that what I write is historically upside-down, since Hilbert first proved what I am claiming in the cyclic case and that this was only later translated in cohomological language...