Let $E$ be an elliptic curve over a number field $k$, and for an extension $K/k$ we denote by $E_K$ the base change $E \times_k K$. By fixing an embedding $k \hookrightarrow \mathbb{C}$, the etale fundamental group $\pi_1(\bar{E})$ is the profinite completion of the topological fundamental group $\pi_1(E(\mathbb{C}))$, and thus $\pi_1(\bar{E})$ is abelian.
Any finite quotient $\Gamma$ of $\pi_1(\bar{E})$ is abelian and corresponds to a covering $\bar{Y} \rightarrow \bar{E}$ that is defined over a finite extension $L/k$. Thus we get a torsor $Y \rightarrow E_L$ under an $L$-group scheme $G$ such that $G(\bar{k}) = \Gamma$. By restriction of scalars, we can construct an $E$-torsor under the abelian group $G$. Hence any $E$-torsor is an abelian covering. This implies that finite descent obstruction to rational points on an elliptic curve is the same as finite abelian descent obstruction, i.e., one has $$E(\mathbb{A}_k)^\mathrm{f-desc} = E(\mathbb{A}_k)^\mathrm{f-ab}.$$
Now if we were to remove finitely many closed points of $E$ to obtain the open (dense) curve $X$, it is no longer true that $\pi_1(\bar{X})$ is abelian. In fact, $X$ constructed in this way is anabelian, and thus it does have nonabelian coverings. However, is it possible that the above equality holds for $X$, i.e., if $Y \rightarrow X$ and $Z \rightarrow X$ are torsors under abelian $G$ and nonabelian $H$ respectively, then if an adelic point $(P_v)$ lifts to some twist $Y^\tau$ under $G$ (since $G^\tau \cong G$), it also lifts to some twist $Z^\xi$ under a nonabelian group $H^\xi$?
If not, does anyone know of an explicit example of such an $X$ such that $X(\mathbb{A}_k)^\mathrm{f-desc} \subsetneq X(\mathbb{A}_k)^\mathrm{f-ab}$?
