For smooth knots in $\mathbb R^3$ from the work of Waldhausen [On Irreducible 3-Manifolds Which are Sufficiently Large, Annals of Mathematics (2) 87.1 (1968), 56–88] it follows that the knot group together with the class of meridian and longitude completely determine the isotopy class of knot. I would like to ask whether something similar can be said about loops of smooth knots? In other words, a loop of smooth knots determines a torus in $\mathbb R^3\times S^1$. I wonder whether the fundamental group of the complement of this torus, possibly together with some extra topological information can determine its isotopy class.
I suppose the most direct analogue would be the fundamental group of the torus exterior in $\mathbb R^3 \times S^1$, together with the homomorphism from that group to $\pi_1 S^1$ (by projection onto the $S^1$ factor), and the longitude-meridian pair on the kernel of that homomorphism. You can reconstruct the torus isotopy class from this data, because the kernel of the map to $\pi_1 S^1$ is a knot group, and the monodromy of that homomorphism tells you the automorphism of the knot group. Since knot exteriors are $K(\pi,1)$ spaces this gives you the homotopy-equivalence of the knot exterior. By Waldhausen, you can promote that to the homeomorphism of the knot exterior, thus the loop of embeddings, up to homotopy.
That said, your graphing construction to build the embedding $S^1 \times S^1 \to \mathbb R^3 \times S^1$ can be refined. There is a "Litherland spinning" construction that rather constructs an embedding $S^2 \to \mathbb R^4$. The key idea is to take your loops in the space of embeddings $I \to D^3$ that are fixed linear embeddings on the boundary. But that's a different question.