Suppose that $a,b \in \mathbb{N}$ are both odd and $m \in \mathbb{Z}.$ I want to show that
$$\left(\dfrac{m}{ab}\right) = \left(\dfrac{m}{a}\right) \left(\dfrac{m}{b}\right)$$
Here, we are talking about the Jacobi Symbol, hence why I put the $\ell$ subscript on each term. I have solved the case when
$$\left(\dfrac{ab}{m}\right) = \left(\dfrac{a}{m}\right) \left(\dfrac{b}{m}\right)$$
Here, we can let $m$ = $p_1 p_2...p_r,$ be the prime factorization of $m$ and we can re-write this as
$$\left(\dfrac{ab}{m}\right) = \left(\dfrac{ab}{p_1}\right) \left(\dfrac{ab}{p_2}\right)... \left(\dfrac{ab}{p_r}\right) = \left(\dfrac{a}{p_1}\right) \left(\dfrac{b}{p_1}\right)... \left(\dfrac{a}{p_r}\right) \left(\dfrac{b}{p_r}\right) = \left(\dfrac{a}{p_1}\right) \left(\dfrac{a}{p_2}\right)... \left(\dfrac{a}{p_r}\right) \left(\dfrac{b}{p_1}\right) \left(\dfrac{b}{p_2}\right)...\left(\dfrac{b}{p_r}\right) = \left(\dfrac{a}{m}\right)\left(\dfrac{b}{m}\right)$$
For the original problem, I want to flip the numerator and denominator in a way that I want get to the form of $\left(\dfrac{ab}{m}\right)$ Is there any way of doing this, or does this question require a completely different proof?
As suggested in the comments, this is an easy consequence of the definition of the Jacobi symbol. But remember that the Jacobi symbol is defined as the product of Legendre symbols, so each 'denominator' needs to be prime.
So, let's suppose $a = p_1 \dotsm p_r$ and $b = q_1 \dotsm q_s$, for each $p_i,q_j$ prime. Then, \begin{align} \left(\dfrac{m}{a}\right) \left(\dfrac{m}{b}\right) &= \left(\dfrac{m}{p_1\dotsm p_r}\right) \left(\dfrac{m}{q_1\dotsm q_s}\right) \\ &=\left(\dfrac{m}{p_1}\right)\dotsm \left(\dfrac{m}{p_r}\right) \left(\dfrac{m}{q_1}\right) \dotsm \left(\dfrac{m}{q_s}\right) \\&=\left(\dfrac{m}{p_1 \dotsm p_r q_1\dotsm q_s}\right) \\ &= \left(\dfrac{m}{ab}\right) \end{align}