Let $M^3$ be a compact $3$-manifold with $\partial M=N$ a connected surface. Suppose one has a smooth embedding of $N$ into the interior of $M$ and $N$ bounds a domain $D$ in $M$. Can we show that $D$ is homeomorphic to $M$?
No, this need not hold.
Here is a concrete example. Suppose that $K \subset S^3$ is the trefoil knot in the three-sphere. It is an old result (certainly known to Alexander) that $K$ is not isotopic to the unknot. Let $N(K)$ be a small tubular neighbourhood of $K$, taken in $S^3$. So $N(K)$ is a solid torus. Let $n(K)$ be the interior of $N(K)$. We define $M = S^3 - n(K)$. This is a (compact) knot exterior. By the "old result" $M$ is not homeomorphic to a solid torus. Note that $T = \partial M = \partial N(K)$ is a two-torus.
Now choose an embedded loop $\alpha$ in $M$. Then $N(\alpha)$, a small tubular neighbourhood of $\alpha$, is another solid torus. Again $T' = \partial N(\alpha)$ is a two-torus. However, $N(\alpha)$ is not homeomorphic to $M$.
The comments above give many other examples. In general, the ways that a fixed surface embeds into a fixed three-manifold is an interesting question.