Product of cosets of normal subgroup and well-definedness Let $G$ be any group and $N$ be a subgroup.
It is proved in almost all the texts of Algebra that, if $N$ is normal, then the product
$$(aN)(bN)=abN$$
is well-defined, i.e. if $aN=a'N$ and $bN=b'N$, then $abN=a'b'N$. But, I confused after some thinking.

If $N$ is normal, then $Nb=bN$, hence
$$(aN)(bN)=a(Nb)N=a(bN)N=abN,$$
and I did not visit in these arguments the question of well-defined-ness!

Q. If we proceed for the product of cosets by above arguments, there is no question of well-defined-ness of the product, am I right?
 A: 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 $(aN)(bN)=abN$ is well defined" and then you say it is not an issue because of "$(aN)(bN)=abN$" 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, $A,B\subseteq G$ some subsets. We define the product of $A$ and $B$ by $AB:=\{ab\ |\ a\in A,\ b\in B\}$.
What you've correctly shown, is that given a normal subgroup $N$ and $a,b\in G$ we have $(aN)(bN)=abN$ with the product of subsets on the left side. This is clear.
Now consider $N\subseteq G$ a normal subgroup. We want to define a multiplication on the set of cosets $G/N$. Meaning we want to define a $M:G/N\times G/N\to G/N$ function (and later on prove something about it). How can we approach this?
Well, the first approach is to define $M(aN,bN):=abN$. Then we have to show that the definition does not depend on the choice of representatives. Which means if $aN=a'N$ and $bN=b'N$ then $abN=a'b'N$. Which is a straightforward application of "$abN=(aN)(bN)$" rule:
$$a'b'N=(a'N)(b'N)=(aN)(bN)=abN$$
Another approach is to define $M(X,Y):=XY$. 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 $X,Y\in G/N$ we have $XY\in G/N$. 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 "$abN=(aN)(bN)$" rule for normal subgroups. Also note that in the previous approach this problem was nonexistent by definition.
So the "$(aN)(bN)=abN$" 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.
A: The answer here is already part of answer by @freakish; but I also found it in the book Elements of Modern Algebra, by Linda Gilbert and Jimmie Gilbert (Section 4.21)
Given any group $G$, one can define product of subsets of $G$ in natural way:
$$S\cdot T=\{st \,\,:\,\, s\in S, \,\, t\in T\}.$$
The associativity property holds for set-products.
When $N$ is a normal subgroup, the product of special subsets is interesting: the product $(aN)\cdot (bN)$ 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 $G/N$ (operation $\cdot$) which does not require well-defined-ness, i.e. it does not need to prove that if $(aN)=(a'N)$ and $(bN)=(b'N)$ then $(aN)\cdot (bN)=(a'N)\cdot (b'N)$ i.e. $abN=a'b'N$.
