Product of cosets of normal subgroup and well-definedness Let G be

Step 1
In order to avoid confusion, lets write down things as clearly and correctly as (hopefuly) needed. The confusion I want to avoid is what follows: you write ''then the product
${\displaystyle \left(aN\right)\left(bN\right)=abN}$
is well defined'' and then you say it is not an issue because of ${\displaystyle \left(aN\right)\left(bN\right)=abN}$ rule. So the question is: is the multiplication in both cases the same thing? The answer is no, or at least: it may be different and only later shown to coincide. In the first sentence it is most likely to be a definition, while in the second a property of some different multiplication, as I will explain now.
Let G be a group, ${\displaystyle A,B\subset qG}$ some subsets. We define the product of A and B by
${\displaystyle AB=\{ab\mid a\in A,b\in B\}}$.
What you've correctly shown, is that given a normal subgroup N and ${\displaystyle a,b\in G}$ we have
${\displaystyle \left(aN\right)\left(bN\right)=abN}$
with the product of subsets on the left side. This is clear.
Now consider ${\displaystyle N\subset qG}$ a normal subgroup. We want to define a multiplication on the set of cosets ${\displaystyle \frac{G}{N}}$. Meaning we want to define a
${\displaystyle M:\frac{G}{N}\times \frac{G}{N}\to \frac{G}{N}}$
function (and later on prove something about it). How can we approach this?
Well, the first approach is to define
${\displaystyle M(aN,bN)=abN}$.
Then we have to show that the definition does not depend on the choice of representatives. Which means if ${\displaystyle aN=a\prime N}$ and ${\displaystyle bN=b\prime N}$ then ${\displaystyle abN=a\prime b\prime N}$. Which is a straightforward application of ${\displaystyle abN=\left(aN\right)\left(bN\right)}$ rule:
${\displaystyle a\prime b\prime N=\left(a\prime N\right)\left(b\prime N\right)=\left(aN\right)\left(bN\right)=abN}$
Another approach is to define ${\displaystyle M(X,Y)=XY}$. Note that this definition doesn't depend on representatives: this form of being well defined is no longer an issue. So we don't have any issues?
Unfortunately, we still need to show that it is well defined, but this time, what it means is that given ${\displaystyle X,Y\in \frac{G}{N}}$ we have ${\displaystyle XY\in \frac{G}{N}}$. It's a valid question, since if N is not normal, then this is not necessarily true. But in this case it again follows from our ''
${\displaystyle abN=\left(aN\right)\left(bN\right)}$
rule for normal subgroups. Also note that in the previous approach this problem was nonexistent by definition.
So the ${\displaystyle \left(aN\right)\left(bN\right)=abN}$ observation is crucial, and it does imply that both approaches are correct (and equal), but it is not really automatic. Also note that these two approaches are very similar and authors may use them interchangeably.

Step 1
Given any group G, one can define product of subsets of G in natural way:
${\displaystyle S\times T=\{st:s\in S,t\in T\}}$.
The associativity property holds for set-products.
When N is a normal subgroup, the product of special subsets is interesting: the product ${\displaystyle \left(aN\right)\times \left(bN\right)}$ of any two left cosets of N can be proved to be a left coset of N, and it is equal to abN; this gives a group structure on ${\displaystyle \frac{G}{N}}$ (operation ${\displaystyle \times }$) which does not require well-defined-ness, i.e. it does not need to prove that if ${\displaystyle \left(aN\right)=\left(a\prime N\right)}$ and ${\displaystyle \left(bN\right)=\left(b\prime N\right)}$ then ${\displaystyle \left(aN\right)\times \left(bN\right)=\left(a\prime N\right)\times \left(b\prime N\right)}$ i.e. ${\displaystyle abN=a\prime b\prime N}$.