The condition that $i_*$ is trivial is neither necessary nor sufficient to to guarantee that $i(\mathbb{T}^2)$ bounds a solid torus.
To show that it is not necessary, consider the manifold $M = S^2 \times S^1$. If $E$ is the equator of $S^2$, then $E\times S^1$ is a solid torus in $M$. However, $\pi_1(M) = \mathbb{Z}$, and the homomorphism $i_*\colon \pi_1(E\times S^1)\to\pi_1(M)$ is onto.
To show that it is not sufficient, consider a standard torus in $S^3$, whose removal separates $S^3$ into two solid tori. Now suppose that we modify one of these solid tori by removing two solid balls and gluing in a copy of $S^2\times[0,1]$ (i.e. by attaching a 3-dimensional "handle"). This will not affect the fact that $i_*$ is trivial, but the inside is no longer a solid torus. If we do the same to the outside, we obtain a torus in a 3-manifold with trivial $i_*$ that does not bound a solid torus.
Indeed, the requirement that $i_*$ is trivial does not even guarantee that the torus is a boundary. For example, if we start with a standard torus in $S^3$ and attach an $S^2\times[0,1]$ handle between the inside and the outside, then $i_*$ will still be trivial, but the torus is no longer the boundary of any region.