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.
The difference in the two cases is that $5\mathbb{Z}$ is a normal subgroup of $\mathbb{Z}$, while $H$ is not normal in $S_3$.
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.
This is in fact, not a group. We can just check via brute force that it has no identity element:
$0$ is not the identity, since $3+_30=0\not=3$.
$1$ is not the identity, since $3+_31=1\not=3$.
$2$ is not the identity, since $3+_32=2\not=3$.
$3$ is not the identity, since $3+_33=0\not=3$.
$4$ is not the identity, since $3+_34=1\not=3$.
$5$ is not the identity, since $3+_35=2\not=3$.
Remember, for an element $e$ to be the identity, we would need that $x+_3e=x$ for every element $x$. But the above computations show that no such $e$ exists here.
Best Answer
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).