Your middle function is not continuous. Let $x_0=0.1$, $y_n=0.4\overbrace{9\cdots9}^{n\text{ times }}0\cdots$. Then $(x_0,y_n)\to(0.1, 0.5)$. We have
$$
f(0.1,0.5)=0.15,\ \ f(0.1,y_n)=0.14090909\cdots
$$
So
$$
|f(0.1,0.5)-f(0.1,y_n)|>0.009
$$
for all $n$.
As a general comment, every time your proof includes a "waving hand part", you should suspect that part. A lot. (said from extensive own experience)
Think about how the plane minus a point looks: it is a disjoint union of concentric circles, each circle being of some radius $r \in (0,\infty)$. (This is how polar coordinates work.)
This essentially is the same as saying that $\mathbb R^2 \setminus \{(0,0)\} \cong S^1 \times (0,\infty).$ Now you just have to remember that $(0,\infty) \cong \mathbb R$.
Rather than asking about general methods of finding homeomorphisms (which is a very difficult problem when stated in full generally), based on what you have written, I think that perhaps you should practice visualizing different examples of products. E.g. you might want to convince yourself that a torus (the surface of a donut) is homeomorphic to $S^1 \times S^1$, by thinking of it as a circle of cirles (just as $\mathbb R^2\setminus \{(0,0)\}$ is a circle of intervals). You could also consider the complement of a ball of radius $1$ in the ball of radius $2$, and convince yourself that it is homeomorphic to the product of $S^2$ and a closed interval.
One important "background" fact to keep in mind is that all open intervals are homeomorphic, and that all closed intervals are homeomorphic, so that the particular endpoints don't matter. (In different concrete situations, it can be helpful to think in terms of different endpoints. E.g. $\mathbb R = (-\infty,\infty)$, but in my explanation of your problem above, it was more convenient visually to think in terms of the open interval $(0,\infty)$.)
Best Answer
The inverse is of course $(s,r)\mapsto rs$. Note that $\frac x{\|x\|}\cdot \|x\|=x$.