This subgroup $A$ is precisely the torsion subgroup in $H^2_{S^1}(M)$ (since $H^2_{S^1}(M^{S^1})$ is obviously free, and the relative $H^1_{S^1}(M,M^{S^1})$ is torsion). So, it will be trivial if and only if the equivariant cohomology in degree 2 is free. This is true for compact $M$ with Hamiltonian $S^1$-action by Kirwan (see, for example, Theorem 14.1 of http://www.math.ias.edu/~goresky/pdf/equivariant.jour.pdf), but I think this is too much to hope for in general (I don't know a connected counterxample off-hand; obviously, you can take a free action on a compact manifold, and take disjoint union with a fixed point).

This subgroup $A$ is precisely the torsion subgroup in $H^2_{S^1}(M)$ (since $H^2_{S^1}(M^{S^1})$ is obviously free, and the relative $H^2_{S^1}(M,M^{S^1})$ is torsion). So, it will be trivial if and only if the equivariant cohomology in degree 2 is free. This is true for compact $M$ with Hamiltonian $S^1$-action by Kirwan (see, for example, Theorem 14.1 of http://www.math.ias.edu/~goresky/pdf/equivariant.jour.pdf), but I think this is too much to hope for in general (I don't know a connected counterxample off-hand; obviously, you can take a free action on a compact manifold, and take disjoint union with a fixed point).