There is already a very similar (basically exact) question posted here, but it shows the opposite direction and I would like to know how to prove it by defining $f: \mathbb{R}^\ast \to \mathbb{R}_{>0} \times \mathbb{Z}_2$ by $f(r) = (|r|, \frac{|r|}{r})$. I've already shown that this is a bijection by showing that $f^{-1}(s, k) = ks$ is both a left and right inverse, but I'm struggling to show that it is a homomorphism. I'm not sure what the group operation on $\mathbb{R}_{>0} \times \mathbb{Z}_2$ is, but I know that the group operation for both $\mathbb{R}_{>0}$ and $\{\pm 1\} \cong \mathbb{Z}_2$ is multiplication. Thus, I assume we're trying to show that $f(xy) = f(x) \cdot f(y)$ for all $x,y \in \mathbb{R}^\ast$. However, I don't really understand the notion of multiplying ordered pairs, as $f(x)$ and $f(y)$ are ordered pairs. Regardless, assuming one just multiplies them element-wise, we have $$f(xy) = \left(|xy|, \frac{xy}{|xy|}\right) = \left(|x||y|, \frac{xy}{|x||y|}\right) = \left(|x|, \frac{x}{|x|}\right) \left(|y|, \frac{y}{|y|}\right) = f(x)f(y)$$
Is this an ok solution? If so, why are we able to multiply ordered pairs element-wise?
Technically, when writing a group, you should write the operation as well. Only when it's abundantly clear what the operation is (either because you've specified it earlier, or because it's a standard group with a single standard operation) can you skip that.
What you've written all looks good.
We are able to multiply ordered pairs element-wise because that's the definition of the group operation in a product group. You should show that what we get with that operation is indeed a group, if you haven't already.