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've got the main idea, so I'll just explain how you might make it rigorous. Suppose you have a presentation $G=\langle g_1,\dots,g_n \mid r_1,\dots,r_m\rangle$ and want to compute the abelianization. First, a group presentation is the same as writing a group as sequence
$$F_m \xrightarrow{\phi} F_n \to G \to 1$$
where $F_m$ and $F_n$ are free groups of orders $m$ and $n$, respectively, with the property that $F_n$ surjects onto $G$ and the normal closure of the image of $F_m$ is the kernel of $F_n\to G$. If we write $x_1,\dots,x_m$ for the generators of $F_m$, then we define $\phi(x_i)=r_i$. Now, we want to compute the abelianization. The abelianization functor is right exact, so applying it to this sequence we get a sequence
$$\mathbb{Z}^m \xrightarrow{\phi_*} \mathbb{Z}^n \to \operatorname{Ab}G \to 0$$
that one can check is exact, where if $e_i=(0,\dots,0,1,0,\dots,0)\in\mathbb{Z}^m$ with a $1$ in position $i$, then $\phi_*(e_i)$ is a vector that gives a signed count of how many times each generator appears in $\phi(x_i)$. For example, if $r_2=g_1g_2g_1g_2^{-1}$ and $n=3$, then $\phi_*(e_2)=(2,0,0)$.
In particular, $\phi_*$ can be written as an $n\times m$ matrix with values in $\mathbb{Z}$, and then we get that $\operatorname{Ab}G$ is the cokernel of this matrix.
This next bit is essentially using the structure theorem for modules over a PID. A useful tool is Smith Normal Form: there exist automorphisms of $\mathbb{Z}^m$ and $\mathbb{Z}^n$ such that the matrix for $\phi_*$ is diagonal and each diagonal entry is divisible by the next. Once it is in this form, the number of all-zero rows gives you the number of $\mathbb{Z}$ factors. If $n\geq m+2$, then there must be at least two $\mathbb{Z}$ factors. As you pointed out, knot groups have only one $\mathbb{Z}$ factor. To tie these together, we need to say that the number of $\mathbb{Z}$ factors is well-defined.
One trick is to tensor everything with $\mathbb{Q}$. Tensoring is right exact, so we get a presentation for $\mathbb{Q}\otimes \operatorname{Ab}(G)$:
$$\mathbb{Q}^m \xrightarrow{\phi_{**}} \mathbb{Q}^n \to \mathbb{Q}\otimes \operatorname{Ab}(G) \to 0$$
The matrix for this presentation is given by normalizing all the nonzero entries on the diagonal to $1$ (since we can divide in $\mathbb{Q}$!). Thus, the dimension of $\mathbb{Q}\otimes \operatorname{Ab}(G)$ as a vector space over $\mathbb{Q}$ is the number of nonzero rows in the matrix, and since dimension is well defined, so is the number of $\mathbb{Z}$ factors.
Best Answer
I haven't thought this through carefully, but just following the hint:
choose a point $a$ on the outer cirlce and a point $b$ on the inner circle. Join $a$ and $b$ with an arc that doesn't cross the two circles. Now "thicken up" this arc very slightly (and thicken up each of $a$ and $b$ to a very short piece of circular arc on their respective circles) so that now we have to the two circles with a very thing rectangle sitting between them. (I think that making this "thickened" picture rigorous is probably part of what is involved in translating the hint into a full argument.)
Now the consider the boundary of the union of the two circles and this rectangle. If you are picturing what I want you to picture, you will see that it is a circle. (Just to be sure: if you take an annulus and snip out a small rectangle joining the inner and outer annulus, you get a kind of "sliced" annulus which is homeomorphic to a disk with circle as boundary. But you can verify that the boundary of the region I am describing is a circle without knowing that the region inbetween the two circles is an annulus just follow the outer circle around from the top left corner of the rectangle all the way around to the top right corner, then move along the right boundary of the rectangle to its bottom right corner, now go back around the inner circle to the bottom left corner of the rectangle, and now go back up the left edge of the rectangle to get back where you started: you've described a circle.) So it bounds a disk. So the region between the two circles, with a thin rectangle removed, is a disk. Gluing back in the rectangle should make into an annulus, as required.