Proposition 3 is false, or at least phrased in a misleading way: the complement of a knotted solid torus is S³ is certainly not a solid torus: this happens if and only if the solid torus is unknotted, essentially because of your argument.
Actually, more is true: a theorem of Gordon and Luecke says that the homeomorphism type of the complement of a knotted solid torus determines the type of the knot.
What's true is that the complement of a standard, revolution solid torus in S³ (which certainly is unknotted) is a solid torus: this doesn't apply to knoted solid tori.
From the perspective of geometrization, satellite knots are knots with a nontrivial JSJ decomposition. This means that there exists an incompressible torus $T$ in the exterior $X$ of the knot $K$ that isn't boundary parallel (i.e., it's not isotopic to the boundary of the knot exterior).
Since $T$ is a torus, a consequence of Dehn's lemma is that $S^3-T$ consists of two components: a solid torus and a knot complement. If $K$ were contained in the knot complement component, then that would mean $T$ bounds a solid torus disjoint from $K$. This contradicts the fact $T$ is incompressible since any meridian disk of the solid torus would serve as a compression disk for $T$. Hence, $K$ is in a solid torus component of $S^3-T$.
If $K$ is contained in a 3-ball in the solid torus component, then one can find a compression disk for $T$ on that side, hence the "geometrically essential knot" condition. Conversely, if there's a compression disk on the solid torus side, one can take the complement of it in the solid torus to get a 3-ball that $K$ is contained within.
If $K$ is a longitude of the solid torus component, then $T$ is boundary parallel, and conversely boundary parallel implies $K$ is a longitude for the solid torus.
If the longitude of the solid torus is an unknot, then there exists a compression disk for $T$ outside the solid torus component (from the disk that the unknot bounds). Conversely, if there's a disk out there then the longitude of the solid torus is an unknot.
Since compression disks are either on one side of $T$ or the other due to orientability of $T$, the three conditions you mention are equivalent to the conditions that $T$ be an incompressible torus that isn't boundary parallel. So, the three conditions you mention are all you need for the definition of a satellite knot for Thurston's classification.
To see torus knots aren't satellite knots, the idea is that the exterior of a torus knot can be given the structure of a Seifert-fibered space. Incompressible non-boundary-parallel tori have to be isotopic to either horizontal surfaces (transverse to each fibers it intersects) or vertical surfaces (unions of fibers). Since $X$ has nonempty boundary, the torus must be a vertical surface, but every vertical torus for a torus knot must either be boundary parallel or bound a solid torus in $X$ due to the fact that the base orbifold is a disk with only two cone points.
Best Answer
(I'm going to use $B^n$ to denote an $n$-dimensional open ball and $D^n$ to denote a closed one.)
Theorem. Let $K$ be a knot. $S^3-\nu(K)\approx S^1\times D^2$ if and only if $K$ is the unknot.
Proof. We will show the stronger statement that $\pi_1(S^3-\nu(K))\approx\mathbb{Z}$ if and only if $K$ is the unknot. Consider a minimal genus Seifert surface $\Sigma$ for $K$, and let $\Sigma'=\Sigma-\nu(K)$. The induced map $\pi_1(\Sigma')\to \pi_1(S^3-\nu(K))$ must be injective, since otherwise by Kneser's lemma there would be a way to compress $\Sigma'$ and reduce its genus. If the knot genus is greater than $0$, then $\pi_1(S^3-\nu(K))$ contains a free group on at least two generators. The rest follows from the fact that the unknot is the unique genus-$0$ knot. $\square$
In fact, Gordon and Luecke proved that if there is an orientation-preserving homeomorphism $S^3-\nu(K)\approx S^3-\nu(K')$, then $K$ and $K'$ are equivalent knots.
Since disks are contractible, the tubular neighborhood $\nu(D)$ of a disk $D$ in $D^4$, regarded as the embedded normal bundle, must be a trivial $\mathbb{R}^2$ bundle over $D$. So, $\nu(D)\approx D\times B^2$. Its closure is $\overline{\nu(D)}\approx D^4$.
The complement $D^4-\nu(D)$ can be rather complicated. While I don't have any examples on hand, it seems that $\pi_1(D^4-\nu(D))\neq \mathbb{Z}$ if $D$ is a ribbon disk for a non-trivial ribbon knot by the van Kampen theorem.
In contrast to the Gordon and Luecke result, I just found a paper by Abe and Tange, "Ribbon disks with the same exterior", where they give a family of inequivalent slice knots such the the complements of a slice disk for each are all diffeomorphic.