This is indeed a duplicate. Just note that the map $S^1 \cong A \to S^1 \times D^2$ induces a map
$\pi_1 S^1 \to \pi_1 (S^1 \times D^2)$
which is the zero map, $0: \mathbb{Z} \to \mathbb{Z}$. This is because $A$ can be shrunk to a point in $S^1 \times D^2$. (You're allowed to homotope $A$ through itself, meaning you can `unlink' it from itself.)
A retraction $S^1 \times D^2 \to A$, however, would induce an isomorphism $\pi_1 A \cong \pi_1 A$ via the composition
$A \to S^1 \times D^2 \to A$
which is impossible because the first map induces the zero map on $\pi_1$.
- I would leave this as a comment but I don't have enough reputation. Matt, it is not true that $\pi_1 A \cong \pi_1(\ast)$. You mean to say that the inclusion $A \to S^1 \times D^2$ induces the zero map, but this does not mean that $\pi_1 A$ is trivial.