In general, mathematicians use the phrase "well defined" when a definition is written in a form that depends (or, rather, seems to depend) on some more or less arbitrary choice. If you make such a definition, you are obligated to show that another choice that satisfied appropriate conditions would lead to the same result.
In a group the product $abc$ is well defined to be $(ab)c$ because its value does not depend on your choice of where to put the parentheses: associativity guarantees $(ab)c = a(bc)$. This fact is so intuitively clear that it's often not made explicit in a beginning algebra course.
When considering quotient groups, you want to define the multiplication of two cosets by choosing an element from each, multiplying them together, and taking the coset of the product. This coset product will be well defined only when the coset of the product of the two group elements does not depend on which ones you happened to choose. The sum of any two odd numbers will be even, so the product of cosets
$(2\mathbb{Z} + 1) \circ (2\mathbb{Z} + 1)$ is $2\mathbb{Z}$.
There is an alternative definition. You can define the product of two cosets $A$ and $B$ as
$$
A \circ B = \{ ab \ | \ a \in A \text{ and } b \in B \}.
$$
This definition does not make any arbitrary choices, but you don't know that the set so defined is really a coset until you prove it.
Write out the $6$ by $6$ multiplication table to verify that $1$ is the identity element, each element has an inverse, and closure is satisfied.
Computing the powers of $3$, and one obtains:
$3^1=3$
$3^2=9$
$3^3= 13$ (because $27$ mod $14$ = $13$)
$3^4 = 11$ (because $3^4 = 3^3 \cdot 3 = 13 \cdot 3 = 39 = 11$)
$3^5 = 5$
$3^6 = 1$.
Thus, the powers of $3$ generate the entire group, whence the group is cyclic.
The map $3^i \rightarrow i$ from the powers of the generator of the group to the powers of the generator of $\mathbb{Z}_6$ gives an isomorphism from the group to $\mathbb{Z}_6$.
The cyclic subgroups of $\mathbb{Z}_n$ are exactly the cyclic groups generated by $d$ for each divisor $d$ of $n$. The cyclic subgroups of $A$ are therefore $\langle 3 \rangle$, $\langle 3^2 \rangle$, $\langle 3^3 \rangle$, and $\langle 3^6 \rangle$.
The given group and $S_3$ are not isomorphic because a cyclic group is abelian and $S_3$ is non-abelian (or, you can say that $S_3$ is non-cyclic whereas the given group is cyclic).
Best Answer
The issue with working with quotient groups is that there are many representatives of the same coset. For example, in $\mathbb{Z}/5\mathbb{Z}$ one has that $$1+5\mathbb{Z}=\{\ldots,-9,-4,1,6,11,\ldots\}=6+5\mathbb{Z}$$ and $$2+5\mathbb{Z}=\{\ldots,-8,-3,2,7,12,\ldots\}=12+5\mathbb{Z}.$$ It is, of course, reasonable to be concerned whether $$3+5\mathbb{Z}=(1+5\mathbb{Z})+(2+5\mathbb{Z})=(6+5\mathbb{Z})+(12+5\mathbb{Z})=18+5\mathbb{Z}?$$ Of course, in this case everything works out just fine, but it is not always so. For example, take the subgroup $H=\langle(12)\rangle=\{(1),(12)\}\leq S_3$. We have $$(13)H=\{(13),(123)\}=(123)H$$ and $$(23)H=\{(23),(321)\}=(321)H$$ However, $(13)(23)H=(321)H$, while $(123)(321)H=(1)H=H$. Hence, $$(13)H(23)H=(13)(23)H=(321)H\neq H=(123)(321)H=(123)H(321)H$$ and the operation is not well defined.
If we assume $H$ is a normal subgroup of $G$, we can show that the operation $aHbH=abH$ is well defined as follows:
Suppose $aH=cH$ and $bH=dH$. By definition, this means that $c^{-1}a\in H$ and $d^{-1}b\in H$. To show that $abH=cdH$, we need to show that $(cd)^{-1}(ab)\in H$.
Well, $$ (cd)^{-1}ab=d^{-1}c^{-1}ab=(d^{-1}(c^{-1}a)d)(d^{-1}b). $$ By assumption $d^{-1}b\in H$. Also, since $c^{-1}a\in H$ and $H$ is normal $d^{-1}(c^{-1}a)d\in H$. Finally, $H$ is a subgroup, so $(d^{-1}(c^{-1}a)d)(d^{-1}b)\in H$ and we're done.