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.
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.
Best Answer
Great question! The issue is:
Below I've used "$+$" instead of "$\#$," to emphasize that the concerns in each case are identical and the only difference is how they're resolved - with inverses not existing in the knot context, and with infinite sums behaving badly in the arithmetic context.
The key point is that in the context of knots, there is a good way to define arbitary infinite sums - where "good" here means that it has nice algebraic properties, and in particular allows the Mazur swindle to go through. It's a bit messy to write down the definition of the infinite connected sum precisely. The simplest approach is to think of knots as continuous injections from $[0,1]$ to the closed unit cube which sends $0$ to $(0,0,0)$ and $1$ to $(1,1,1)$ (intuitively, the actual knot is formed by joining up these two points) and of equivalence of knots amounting to isotopy fixing those basepoints.
Now, we compose two knots intuitively by putting two unit cubes "corner-to-corner," drawing the respective knots in those cubes, and then "scaling down" by a factor of $2$ in each direction. But we can also compose infinitely many knots! Specifically, we start by putting an infinite chain of cubes corner-to-corner, and placing the corresponding knots in each. We then scale down in a more complicated way, as: $$(a,b,c)\mapsto ({2\arctan a\over \pi},{2\arctan b\over\pi}, {2\arctan c\over\pi}).$$ This fits our infinite concatenation of (shifted) knots into the unit cube; we then add the point $(1,1,1)$ to the whole shebang to get a genuine knot.
The key point is that this is totally well-defined. (Well, it would be if written out a bit more formally, but meh.) Our next step is to rigorously prove things about it; specifically, that it satisfies the appropriate "infinitary associativity law." This isn't hard to do - the isotopy in question is quite easy to write down, if a bit tedious. What we then get out of this is that, for any sequence of knots $(K_i)_{i\in\mathbb{N}}$, the infinite sums $$\sum_{i\in\mathbb{N}}(K_{2i}+K_{2i+1})$$ and $$K_0+\sum_{i\in\mathbb{N}}(K_{2i+1}+K_{2i+2})$$ are each defined and are equal (fine, they're not literally the same knot, but they represent the same knot class). And from this, we get the Mazur swindle.
Looking at this we can see where the analogous "proof" for arithmetic is incomplete: to finish it, we would need to $(i)$ find a way to assign a real number to every expression of the form $\sum_{i\in\mathbb{N}}x_i$, and then $(ii)$ show that that assignment satisfied the appropriate infinitary associativity law. Certainly the usual definition via the limit of an infinite sequence doesn't help us here, since it's not always defined (in particular, $\lim_{n\rightarrow\infty}(\sum_{i=1}^n (-1)^i$ does not exist).
Indeed what we learn from the usual paradox is that this can't be done.$^1$ More broadly, we get the general theorem that - roughly speaking - we can never have a context where all infinite sums make sense and behave well, every element has an inverse, and not everything is equal to zero. I don't think this result has a specific name; I've heard it referred to as the Eilenberg-Mazur swindle as well, since it's an immediate fallout of that (if I recall correctly, Eilenberg introduced the same argument in an algebraic - as opposed to geometric - context, at around the same time as Mazur introduced it in knot theory).
$^1$That said, there's lots of interesting math around partial definitions along these lines - that is, notions of "infinite sum" which $(i)$ extend the usual notion to at least some additional series and $(ii)$ have some basic niceness properties. See e.g. here.