If an orientable 3-manifold $M$ has boundary the torus $S^1\times S^1$ and deform retracts to a solid torus $S^1\times D^2$, is it necessarily homeomorphic to a solid torus?
Equivalently, if the complement of a knot in a closed orientable 3-manifold deform retracts to a circle, is the knot complement a solid torus?
Yes, this is true.
First, since $\partial M \hookrightarrow M$ is not $\pi_1$-injective, by the Loop Theorem there is a properly embedded 2-disc $(D,\partial D) \hookrightarrow (M,\partial M)$ such that $\partial D$ does not bound a disc in the torus $\partial M$.
Second, letting $N$ be a regular neighborhood of $\partial M \cup D$, the surface $\partial N$ is a separating 2-sphere $S \subset \text{interior}(M)$. Let the closures of the components of $M-S$ be $M_1,M_2$ where $\partial M \subset M_1$.
Third, by hypothesis the inclusion $M_1 \hookrightarrow M$ is injective on $\pi_1$. Therefore by Van Kampen's theorem, $M_2$ is simply connected.
Fourth, by the Poincare Conjecture, now Perelman's Theorem, $M_2$ is a 3-ball (before Perelman one used to throw in the hypothesis "assume $M$ contains no fake 3-ball").
It follows that $M$ is a solid torus.