I'll be working in $S^3$ for this answer, as it is generally the more convenient place to do knot theory. It does not cause problems: just extend your homeomorphisms to be homeomorphisms of $S^3$, for instance.
These are not identical, but they almost are. In particular, the reverse of a knot need not be isotopic to it, but you can just reflect along a hyperplane to get a homeomorphism.
But if you replace "homeomorphism" with "orientation-preserving homeomorphism", they are equivalent. This is because every orientation-preserving homeomorphism of $S^3$ is isotopic to the identity! This follows from a combination of Alexander's trick and the isotopy extension theorem. If you want to do knot theory in the smooth category, working with smooth isotopies instead of continuous isotopies, you need to know that every orientation-preserving diffeomorphism is isotopic to the identity; I think Cerf was the first to prove this.
There is one other common definition of equivalence for knots: isotopy of embeddings.
A knot is an embedding $S^1 \to S^3$. An isotopy between two embeddings $f_0, f_1$ is a map $f_t: S^1 \times [0,1] \to S^3$ such that each $f_t$ is an embedding. This turns out to not be the correct notion of equivalence for knots - it would force all tame knots to be equivalent! What one does to fix this is either demand that the isotopy be smooth, or that it be locally flat. (See this page again for details on being locally flat.)
Now this is precisely the hypothesis we need to use the isotopy extension theorem: so any (locally flat/smooth) isotopy of embeddings automatically lifts to an ambient isotopy, giving us something equivalent to the standard definition.
EDIT: I was having some trouble reconstructing the argument I suggested above to show that $\pi_0 \text{Homeo}^+(S^3)$ is trivial. So I'll write an argument here. Actually I'll show that $\pi_0 \text{Homeo}^+(S^n) = 0$ for all $n$.
This is trivial for $n=0$. Inductively assume it's trivial for $n-1$. Pick an orientation-preserving homeomorphism $f$; $f(S^{n-1})$ bounds a ball on both sides. Isotope $f$ so that $f(S^{n-1})$ does not include the poles. Pick a standard 'longitude sphere' $S^{n-1}$s (the level sets of the height function of $S^n \subset \mathbb R^{n+1}$) that lie inside one of the balls that $f(S^{n-1})$ bounds (this is possible by compactness of $S^{n-1}$ and continuity of the height function). By the annulus theorem there is a locally flat isotopy of embeddings from $f(S^{n-1})$ to this longitude sphere; then there is an isotopy from this longitude sphere to the equator $S^{n-1}$. Using the isotopy extension theorem $f$ then is isotopic to an orientation-preserving homeomorphism that preserves the equator; we can also assume that it sends the north hemisphere to the north hemisphere. This new $f'$ restricts to an orientation-preserving map on $S^{n-1}$; this restriction is isotopic to the identity; use isotopy extension. So $f'$ is isotopic to $f''$ that preserves $S^{n-1}$ pointwise. $f''$ is the union of two homeomorphisms of $D^n$ (the northern and southern hemispheres), identity on the boundary, along their boundary; the Alexander trick shows that these are isotopic to the identity. So we have produced an isotopy of $f$ to the identity as desired.
In the case we care about - $n=3$ - the only hard result we used is the annulus theorem in dimension 3, which is reasonably elementary.
You can pick up some of this theory from the likes of Hatcher's notes on 3-manifolds or Hempel's or Jaco's books on 3-manifold topology, or perhaps even Rolfsen's book on knot theory. Lickorish sprinkles some here and there at various points in his book.
The intuition for a ball $B\subset S^3$ meeting $K$ as a trivial ball-arc pair being "essentially unique" is that you can shrink $B$ down so that it can be thought of as a tiny bead on $K$. Then, any two beads on $K$ are equivalent by sliding them around.
A trivial ball-arc pair $(B,\alpha)$ is a "ball $B$ with an unknotted spanning arc $\alpha\subset B$", which more precisely is a pair that is, as a subspace, PL homeomorphic to the pair $$[0,1]\times (D^2,0)=([0,1]\times D^2,[0,1]\times 0).$$
Let $(B,\alpha)\subset (S^3,K)$ be a trivial ball-arc pair.
By Alexander's theorem (that for a PL embedded $2$-sphere $S\subset S^3$, the closure of each component of $S^3-S$ is PL homeomorphic to a $3$-ball with boundary $S$) and Alexander's trick, we can get an isotopy of $S^3$ that isotopes $B$ to a standard $3$-ball in $S^3$ while isotoping $\alpha$ to a diameter of $B$. Alexander's trick uses the PL homeomorphism that characterizes a trivial ball-arc pair.
Once in this position, we can make a knot diagram for $K$ where $\alpha$ is indicated by small arc in the diagram away from any crossings, where $B$ projects onto a small disk with the arc as its diameter. We are allowed to slide this arc around the diagram (crossing crossings) since there is a corresponding isotopy of $S^3$ to implement this move. Reidemeister moves (which correspond to certain isotopies of $S^3$) are allowed at least if the arc is not in the way, but the arc can always be moved to be out of the way. Also, isotopies of the diagram itself correspond to isotopies of $S^3$ as usual. From here, it should be clear that any trivial ball-arc pair is equivalent to any other, since the Reidemeister moves and isotopies of diagrams generate knot equivalence.
I used the Reidemeister moves as a tool to deal with $S^3$ isotopies, but this was not necessary. Other options include (1) taking the ball-arc pair and extending it into a closed tubular neighborhood of $K$, then using the fact that every parameterization of a closed tubular neighborhood is isotopic to any other, and then finishing it off with the fact that any closed interval on $S^1$ is isotopic to any other, (2) using the parameterization of the trivial ball-arc pair to shrink it so that it lies inside a given tubular neighborhood, then seeing the way in which it sits with respect the foliation of the tubular neighborhood by disks to isotope it (depending on the present singularities) into a standard ball with only a maximum and minimum, and any pair of such balls are more-or-less-obviously isotopic. Most of the complexity is keeping track of the arc inside the ball.
Best Answer
This definition is relying on a key fact about PL (or smooth) topology: if $h: S^3 \to S^3$ is an orientation-preserving PL homeomorphism, then there is an isotopy $H : [0,1]\times S^3\to S^3$ such that $H_0=\operatorname{id}_{S^3}$ and $H_1=h$. This is because the mapping class group of $S^3$ is trivial. Since $h(L_1)=L_2$, then $H_t|_{L_1}:L_1\to S^3$ is an isotopy from $L_1$ to $L_2$ through PL embeddings.
The unrestricted $H$ is known as an ambient isotopy. What you want from a definition of isotopy of knots is isotopy extension to ambient isotopies. Intuitively, dragging the knots around should extend to dragging around the ambient space, too. Why is this? You want any sorts of peripheral structures, like Seifert surfaces, to be able to follow along the isotopy, too. If you have a continuous family $h:[0,1]\times S^1 \to S^3$ of PL embeddings, then this does indeed extend to an ambient isotopy. And since the mapping class group is trivial, the only data you need out of this is the single orientation-preserving PL homeomorphism of $S^3$ that carries the knot to the end result of the isotopy.
There is a strange detail in here: while $h:S^3\to S^3$ does come from an ambient isotopy, there can be many ambient isotopies it comes from that are not isotopic to each other (yes, non-isotopic isotopies :-)). This can happen when a knot is a connect sum: a connect sum of two right-handed trefoil knots has an isotopy that swaps the two connect summands, and this isotopy should be non-isotopic to the identity isotopy. This detail does not matter for the definition of knot equivalence, though.