I am looking for an example of a $3$-dimensional compact and orientable topological manifold $\mathcal{M}$ with boundary, such that $\partial\mathcal{M}$ is homeomorphic to a $2$-torus $T^{2}=S^{1}\times S^{1}$, but such that $\mathcal{M}$ is not homeomorphic to the solid torus $\overline{T}^{2}:=S^{1}\times\overline{D}$, where $\overline{D}$ denotes the closed disk (=closed $2$-ball).
This is basically a similar question as in this post, which I've created some days ago. In this post I asked for manifolds, with spherical boundary, which are not homeomorphic to a $3$-ball. As explained by Moishe Kohan in this post, a general procedure to built them is to consider an arbitrary closed orientable $3$-manifold (which is not equivalent to $S^{3}$), from which we "cut out" the interior of a regular $3$-ball. Is there a similar procedure which works for the torus case?
I already know from the well-known theorem $\chi(\mathcal{M})=\frac{1}{2}\chi(\partial\mathcal{M})$ for odd-dimensional compact manifolds that every manifold $\mathcal{M}$ with toroidal boundary has to have $\chi(\mathcal{M})=0$. So, whenever someone knows an example of a $3$-dimensional compact and orientable manifold with a single boundary component, whose Euler-characterisitc is zero, then we would be done.
Yes, the same procedure works for the torus case: take an arbitrary closed orientable 3-manifold from which we "cut out" the interior of an embedded regular solid torus.
In this situation there is no need to rule out $S^3$. In fact $S^3$ gives us gobs of beautiful examples closely related to knot complements: in fact what this construction shows is that every knot complement is the interior of a 3-manifold with torus boundary.