Let $K \subset S^3$ be a knot. Given $r \in \mathbb{Q} \cup \{ \infty\}$, denote by $S^3_{r}(K)$ the 3-manifold obtained by Dehn surgery on $K$ with coefficient $r$. Is it true that: $S^3_r(K)$ contains a incompressible torus if and only if $K$ is a satellite with non trivial companion?
Recall that Thurston proved that a knot is a satellite with non trivial companion if and only if an essential torus is contained in its complement.
A somewhat belated answer, but never mind.
It turns out that neither direction is true.
For an example of a non satellite knot with a surgery containing an incompressible torus consider consider $S_0^3(K)$, where $K$ is either the figure 8 knot or the trefoil. In either case, the $K$ is fibred of genus one, so the $0$-surgery is $T^2$-bundle over $S^1$. Where a $T^2$ in this bundle is obtained by taking a Seifert surface and capping off the puncture with a core disk of the Dehn filling. This $T^2$ is incompressible; the long exact homotopy sequence of a fibration has terms $$0=\pi_2(S^1) \rightarrow \pi_1(T^2) \rightarrow \pi_1(S_0^3(K)),$$ showing that the torus is $\pi_1$-injective.
For the other direction, one needs a knot $C$ in the solid torus with non-trivial surgeries again yielding a solid torus. Using such a knot as the companion, with a non-satellite pattern $K$, yields a knot $C(K)$ with surgery slopes for which the incompressible torus disappears and for which surgery on $C(K)$ is homeomorphic to surgery on $K$ for some other slope. Such knots in the solid torus exist and, in fact, were classified by Gabai, who showed that they were either torus knots or certain 1-bridge braids (also known as Berge-Gabai knots). A concrete example would be something like $7$-surgery on the $(2,3)$-cable of the trefoil. This will yield a manifold homeomorphic to $7/4$-surgery on the trefoil, which is a small Seifert fibered space and hence contains no incompressible tori.