Let $\frac{a}{b},\frac{c}{d}\in \mathbb{Q}$. prove that the product is well-defined We have that $\frac{a}{b}\cdot \frac{c}{d}=\frac{a'}{b'}\cdot\frac{c'}{d'}$, which can be rewritten as $\frac{ac}{bd} = \frac{a'c'}{b'd'}$ and again as $(ac)(b'd') = (bd)(a'c').$

If $\frac{a}{b} = \frac{a'}{b'}$ and $\frac{c}{d} = \frac{c'}{d'}, $ then $ab' = ba'$ and $cd' = dc'.$

$= (ab')(dd') = (ba')(dd')$ and $(cd')(bb') = (dc')(bb')$

$= (ab')(dd')(cd')(bb') = (ba')(dd')(dc')(bb')$ when multiplying the two equations.

Here is where I got stuck. I was attempting to get the final equation equal to $(ac)(b'd') = (bd)(a'c')$ so as to prove that the multiplication is well-defined, but I am unsure of how to get there from where I am or whether I'm thinking about this wrong to begin with.
You are correct that you need to prove that $(ac)(b'd')=(bd)(a'c')$, but you're making it too complicated. By definition,

$$\frac{a}{b} = \frac{a'}{b'} \: \: \text{and} \:\: \frac{c}{d} = \frac{c'}{d'}, $$

means that $ab' = ba'$ and $cd' = dc'$. By multiplying these two equations together (ie left times left, right times right) you simply get $(ac)(b'd')=(bd)(a'c')$, which is what you're looking for.
You can simply say: 
$$\frac{a}{b}\cdot\frac{c}{d}=\frac{a}{b}\cdot\frac{c}{d}$$
which is always verified.
For the hypothesis done $\left (\frac{a}{b}=\frac{a_1}{b_1}, \frac{c_1}{d_1}=\frac{c}{d}\right)$, we arrive at:
$$\frac{a}{b}\cdot\frac{c}{d}=\frac{a_1}{b_1}\cdot\frac{c_1}{d_1}$$