I knot of no analogue of property P.

In fact, there are many links in $S^3$ which admit infinitely many fillings which are also $S^3$. The simplest is probably the Whitehead link: each component is unknotted, and one can "twist" along the unknotting disk to obtain infinitely many links with the same complement.

The only thing that I know of is a kind of converse to this by Cameron Gordon. If a link has infinitely many Dehn fillings yielding $S^3$, then there is a collection of disjointly embedded disks and annuli in $S^3$ bounding a sublink so that all but finitely many of the fillings are obtained by $1/n$ filling along the boundaries of the disk complements, and pairs of fillings along the boundaries of the annulus components. This is not exactly stated in his paper, but I think that it follows from his proof. Applying to knots, one sees that a knot with infinitely many $S^3$ fillings is the unknot. But this is quite weak compared to property P.

I know of no analogue of property P.

In fact, there are many links in $S^3$ which admit infinitely many fillings which are also $S^3$. The simplest is probably the Whitehead link: each component is unknotted, and one can "twist" along an unknotting disk bounding one component to obtain infinitely many non isotopic links with the same complement.

The only thing that I know of is a kind of converse to this by Cameron Gordon. If a link has infinitely many Dehn fillings yielding $S^3$, then there is a collection of disjointly embedded disks and annuli in $S^3$ bounding a sublink so that all but finitely many of the fillings are obtained by $1/n$ filling along the boundaries of the disk complements, and pairs of fillings along the boundaries of the annulus components. This is not exactly stated in his paper, but I think that it follows from his proof. Applying to knots, one sees that a knot with infinitely many $S^3$ fillings is the unknot. But this is quite weak compared to property P.