This is a question that I remember worrying about when I first started learning about knot theory. Older books have a tendency to skim over this point rather lightly, perhaps because the resolution of the question seems to involve techniques that have little to do with standard knot theory.  One book that doesn't avoid treating this issue squarely is Burde and Zieschang's "Knots" (now in a third edition with a third author, Heusener, and available in an electronic version). In Chapter 1 of this book the fact that every orientation-preserving homeomorphism of $S^3$ (or equivalently ${\mathbb R}^3$) is isotopic to the identity is attributed to a 1960 paper of G.M.Fisher. (In Kirby's landmark 1969 paper on the Stable Homeomorphism Conjecture he refers to this result just as being "classical".) 

The earlier work of Moise on existence and uniqueness of PL structures on 3-manifolds does not seem to say anything about homeomorphisms between PL 3-manifolds being isotopic to PL homeomorphisms. Moise just proves that such homeomorphisms can be approximated by PL homeomorphisms. However, one of his approximation theorems appears to be sufficient to deduce the desired fact about knots. Namely, suppose $f:S^3\to S^3$ is an orientation-preserving homeomorphism taking a knot $K$ to a knot $K'$. Choose an open set $U$ in $S^3-K$ and choose a continuous function $\varepsilon: U \to (0,1)$ that approaches $0$ at the frontier of $U$. Then Moise's theorem states that $f$ can be $\varepsilon$-approximated by a homeomorphism $g$ which is PL on any compact set in $U$ (which just means PL on $U$ itself) and which equals $f$ outside $U$. In particular $g$ takes $K$ to $K'$. Since $g$ is PL on some open set, it can easily be isotoped (topologically) to the identity, and this isotopy restricts to a topological isotopy of $K'$ to $K$. If $K$ and $K'$ are PL knots, one might want a PL isotopy, but this argument doesn't give it.

The Burde-Zieschang book does prove that two PL knots that are equivalent via a topological homeomorphism that preserves orientation are PL isotopic. This is Corollary 3.17, and they deduce it from Waldhausen's theorem that, in the PL category, two knots are isotopic if they have isomorphic "peripheral systems". The latter are defined in terms of the fundamental group of the knot complement, so they are invariant under topological homeomorphisms preserving orientation.

I don't know who first proved the general result that every homeomorphism between PL 3-manifolds is isotopic to a PL homeomorphism, but there is a very nice proof using just PL topology of 3-manifolds and the Kirby torus trick in a 1976 paper of A.J.S.Hamilton.

In the smooth category Cerf's theorem ($\Gamma_4=0$) certainly implies that if two smooth knots are equivalent under an orientation-preserving diffeomorphism then they are smoothly isotopic. However, one can obtain this without using Cerf's theorem by an elementary argument that works in all dimensions, even when the analog of Cerf's theorem fails (due to the existence of exotic spheres). Suppose $f:S^n\to S^n$ is an orientation-preserving diffeomorphism taking a closed proper subset $K\subset S^n$ to another such set $K'$. Since $f$ preserves orientation, it is a standard fact that $f$ can be isotoped to be the identity on a ball $B\subset S^n-K$ staying fixed outside a larger ball in $S^n-K$, so we assume this has been done. Another well-known elementary fact is that the map from the diffeomorphism group $Diff(B,\partial B)$ of $B$ fixing (a neighborhood of) $\partial B$ to the orientation-preserving diffeomorphism group $Diff^+(S^n)$ of $S^n$, obtained by extending diffeomorphisms via the identity outside $B$, is a weak homotopy equivalence. All we need however is the fact that this map induces a surjection on $\pi_0$, which is equivalent to the previously stated standard fact. This means that we can compose the given $f$ with a diffeomorphism supported in $B$ representing the inverse element of $\pi_0Diff^+(S^n)$ to get a new diffeomorphism that is isotopic to the identity and still takes $K$ to $K'$. (This can be done quite explicitly in fact.)