I am new to knot theory. So the following problem description may not be rigorous; but I will try my best to explain intuitively. Please tell me if there is ambiguity.
For any self-intersected curve $S$ on the XY plane, is it always possible to find a corresponding non-self-intersected knot curve in 3D such that its projection on the XY plane is exactly $S$?
In general, no. For example if you have an infinity of self-intersection it is not possible.
But as long as your curve is not too much pathological, yes. If you have a "reasonable" parametrization from $S^1$ to $S$, what you can do is consider the corresponding parametrization $s : [0,1] \rightarrow \mathbb{R}^2$ and create a knot in $\mathbb{R}^3$ by constructing the curve $[0,1] \rightarrow \mathbb{R}^3 : a \rightarrow (S(a).x, S(a).y, a)$, we now just need to close the loop by adding the segment $(S(0).x, S(0).y, [0,1])$ (for that you need $S(0)$ not to be an intersection).
A short remark on your question: it is important to see that a knot is never self-intersecting. (At least if we consider standard knot theory where singular knots does not exist)