As Ryan says, this follows from Waldhausen's paper, when appropriately interpreted. Sufficiently large 3-manifolds are usually called "Haken" in the literature, and as Ryan says, they are irreducible and contain an incompressible surface (which means that the surface is incompressible and boundary incompressible). An irreducible manifold with non-empty boundary and not a ball (ie no 2-sphere boundary components) is always sufficiently large, by a homology and surgery argument. By Alexander's Lemma, knot complements are irreducible, and therefore sufficiently large (the sphere theorem implies that they
are aspherical).
Waldhausen's theorem implies that if one has two sufficiently large 3-manifolds $M_1, M_2$ with connected boundary components, and an isomorphism $\pi_1(M_1) \to \pi_1(M_2)$ inducing an isomorphism $\pi_1(\partial M_1) \to \pi_1(\partial M_2)$, then $M_1$ is homeomorphic to $M_2$. This is proven by first showing that there is a homotopy equivalence $M_1\simeq M_2$ which restricts to a homotopy equivalence $\partial M_1\simeq \partial M_2$. Then Waldhausen shows that this relative homotopy equivalence is homotopic to a homeomorphism by induction on a hierarchy. The peripheral data is necessary if the manifold has essential annuli, for example the square and granny knots have homotopy equivalent complements.
If $K_1, K_2\subset S^3$ are (tame) knots, and $M_1=S^3-\mathcal{N}(K_1), M_2=S^3-\mathcal{N}(K_2)$ are two knot complements, then Waldhausen's theorem applies. However, one must also cite the knot complement problem solved by Gordon and Luecke, in order to conclude that $K_1$ and $K_2$ are isotopic knots. Otherwise, one must also hypothesize that the isomorphism $\partial M_1 \to \partial M_2$ takes the meridian to the meridian (the longitudes are determined homologically). This extra data is necessary to solve the isotopy problem for knots in a general 3-manifold $M$, to guarantee that the homeomorphism $(M_1,\partial M_1)\to (M_2,\partial M_2)$ extends to a homeomorphism $(M,K_1)\to (M,K_2)$, since for example there are knots in lens spaces which have homeomorphic complements by a result of Bleiler-Hodgson-Weeks.
Yes, there's loads of other contractible 4-manifolds bounding homology spheres with other geometries.
For example $1$-surgery on the Stevedore knot is a hyperbolic manifold with volume $1.3985\cdots$. This bounds a contractible manifold by the same kind of arguments Casson and Harer used to cook up their big list of Mazur manifolds. There are other examples like this that occur in my arXiv preprint: https://arxiv.org/abs/0810.2346 Some are geometric, some have incompressible tori.
I don't think any Nil manifolds bound contractible 4-manifolds. Crisp and Hillman determined all the Nil manifolds that embed smoothly (or topologically) in the $4$-sphere. The only Nil manifolds on that list have non-trivial homology. Crisp & Hillman "Embedding Seifert fibred 3-manifolds and Sol-manifolds in 4-space".
Best Answer
Perelman has proved Thurston's geometrization conjecture, which says that every irreducible 3-manifold decomposes along its canonical decomposition along tori into pieces, each admitting a geometric structure. A "geometric structure" is a nice riemannian metric, which is in particular complete and of finite volume.
There are eight geometric structures for 3 manifolds: three structures are the constant curvature ones (spherical, flat, hyperbolic), while the other 5 structures are some kind of mixing of low-dimensional structures (for instance, a surface $\Sigma$ of genus $2$ has a hyperbolic metric, and the three-manifold $\Sigma\times S^1$ has a mixed hyperbolic $\times S^1$ structure).
The funny thing is that geometrization conjecture was already proved by Thurston when the canonical decomposition is non-trivial, i.e. when there is at least one torus in it. In that case the manifold is a Haken manifold because it contains a surface whose fundamental group injects in the 3-manifold. Haken manifolds have been studied by Haken himself (of course) and by Waldhausen, who proved in 1968 that two Haken manifolds with isomorphic fundamental groups are in fact homeomorphic.
If the canonical decomposition of our irreducible manifold $M$ is empty, now we can state by Perelman's work that $M$ admits one of these 8 nice geometries. The manifolds belonging to 7 of these geometries are well-known and have been classified some decades ago (six of these geometries actually coincide with the well-known Seifert manifolds, classified by Seifert already in 1933). From the classification one can see that the only distinct manifolds with isomorphic fundamental groups are lens spaces (which belong to the elliptic geometry, since they have finite fundamental group).
The only un-classified geometry is the hyperbolic one. However, Mostow rigidity theorem says that two hyperbolic manifolds with isomorphic fundamental group are isometric, hence we are done. Some simple considerations also show that two manifolds belonging to distinct geometries have non-isomorphic fundamental groups.
Therefore now we know that the fundamental group is a complete invariant for irreducible 3-manifolds, except lens spaces.