As was pointed out by Felipe Voloch in the comments, this obstruction is equivalent to the finite abelian descent obstruction [Stoll, Thm. 6.4 and the discussion before Cor. 6.2]. It is also shown that for curves, this agrees with the Brauer–Manin obstruction [Stoll, Cor. 7.3]. Finally, Stoll conjectures the following:

Conjecture [Stoll, Conj. 9.1]. Let $C$ be a smooth projective geometrically connected curve over a number field $K$. Then $\overline{C(K)} = C(\mathbb A_K)_\bullet^{\text{f-ab}}$ (see [Stoll, Def. 5.4 and §2] for notation).

That is, the finite abelian descent obstruction is the only obstruction to the Hasse principle and weak approximation.

What is known.

• If $g = 0$, then $C$ satisfies the Hasse principle $C(K) = \varnothing \Leftrightarrow C(\mathbb A_K)_\bullet = \varnothing$. If $C(\mathbb A_K)_\bullet = \varnothing$ then there is nothing to prove, and if $C \cong \mathbb P^1_K$ then $C(K)$ is dense in $C(\mathbb A_K)_\bullet$.
• If $g = 1$, then the conjecture is true if and only if $Ш(K,\operatorname{Pic}^0_C)_{\text{div}} = 0$ [Stoll, Cor. 6.2 and the discussion after Cor. 6.3]. In particular, this is a weak form of the Tate–Shafarevich conjecture.
• If $g \geq 2$, then $C(K)$ is finite by Faltings, so $\overline{C(K)} = C(K)$. Thus, the conjecture says that $C(\mathbb A_K)_\bullet = C(K)$. I'm not even sure if we know whether $C(\mathbb A_K)_\bullet$ is finite.

Finally, Stoll proves the conjecture in the following case:

Theorem [Stoll, Thm. 8.6]. Assume $C$ admits a nonconstant map $C \to A$ into an abelian variety $A$ such that $A(K)$ is finite and $Ш(K,A)_{\text{div}} = 0$. Then $C(K) = C(\mathbb A_K)_\bullet$.

This is for example the case when $K = \mathbb Q$ and $A$ is modular of analytic rank $0$. As mentioned before, the condition $Ш(K,A)_{\text{div}}$ is expected to always be true, but the condition that $A(K)$ is finite is a serious restriction on the generality of the theorem.

Recent progress?

I haven't worked on rational points for a while, so I don't know what the current status is. I would be surprised if there were a great breakthrough on the conjecture in its general form, though.

References.

[Stoll] Stoll, Michael, Finite descent obstructions and rational points on curves, Algebra Number Theory 1, No. 4, 349-391 (2007). ZBL1167.11024.

As was pointed out by Felipe Voloch in the comments, this obstruction is equivalent to the finite abelian descent obstruction [Stoll, Thm. 6.4 and the discussion before Cor. 6.2]. It is also shown that for curves, this agrees with the Brauer–Manin obstruction [Stoll, Cor. 7.3]. Finally, Stoll conjectures the following:

Conjecture [Stoll, Conj. 9.1]. Let $C$ be a smooth projective geometrically connected curve over a number field $K$. Then $\overline{C(K)} = C(\mathbf A_K)_\bullet^{\text{f-ab}}$ (see [Stoll, Def. 5.4 and §2] for notation).

That is, the finite abelian descent obstruction is the only obstruction to the Hasse principle and weak approximation.

What is known.

• If $g = 0$, then $C$ satisfies the Hasse principle $C(K) = \varnothing \Leftrightarrow C(\mathbf A_K)_\bullet = \varnothing$. If $C(\mathbf A_K)_\bullet = \varnothing$ then there is nothing to prove, and if $C \cong \mathbf P^1_K$ then $C(K)$ is dense in $C(\mathbf A_K)_\bullet$.
• If $g = 1$, then the conjecture is true if and only if $Ш(K,\operatorname{Pic}^0_C)_{\text{div}} = 0$ [Stoll, Cor. 6.2 and the discussion after Cor. 6.3]. In particular, this is a weak form of the Tate–Shafarevich conjecture.
• If $g \geq 2$, then $C(K)$ is finite by Faltings, so $\overline{C(K)} = C(K)$. Thus, the conjecture says that $C(\mathbf A_K)_\bullet^{\text{f-ab}} = C(K)$. I'm not even sure if we know whether $C(\mathbf A_K)_\bullet^{\text{f-ab}}$ is finite.

Finally, Stoll proves the conjecture in the following case:

Theorem [Stoll, Thm. 8.6]. Assume $C$ admits a nonconstant map $C \to A$ into an abelian variety $A$ such that $A(K)$ is finite and $Ш(K,A)_{\text{div}} = 0$. Then $C(K) = C(\mathbf A_K)_\bullet^{\text{f-ab}}$.

This is for example the case when $K = \mathbf Q$ and $A$ is modular of analytic rank $0$. As mentioned before, the condition $Ш(K,A)_{\text{div}}$ is expected to always be true, but the condition that $A(K)$ is finite is a serious restriction on the generality of the theorem.

Recent progress?

I haven't worked on rational points for a while, so I don't know what the current status is. I would be surprised if there were a great breakthrough on the conjecture in its general form, though.

References.

[Stoll] Stoll, Michael, Finite descent obstructions and rational points on curves, Algebra Number Theory 1, No. 4, 349-391 (2007). ZBL1167.11024.