Do subgroup and quotient group define a group?

No.

Both $S_3$ and $C_6$ have normal subgroups isomorphic to $C_3$ with quotient isomorphic to $C_2$.

This is not even true of abelian groups: $C_4$ and $C_2 \times C_2$ both have subgroups isomorphic to $C_2$ with quotient isomorphic to $C_2$.

No, your question is concerned with "Extension theory for groups". If you have $N \unlhd G$ a normal subgroup then let $\phi$ be the natural inclusion (injective) $\phi \colon N \hookrightarrow G$ and $N$ is the kernel of the natural map $\pi \colon G \twoheadrightarrow G / N$. Note that $im(\phi)=\phi(N)=N=ker(\pi)$. That is to say that $G$ fits into the sequence $$ 0 \rightarrow N \rightarrow G \rightarrow G /N \rightarrow 0$$ which is exact, i.e. the second arrow is injective (this is $\phi$), the third is surjective (this is $\pi$) and $im(\phi)=ker(\pi)$. One now asks which groups $G$ fit into such an exact sequence, i.e for given groups $\Gamma$ and $Q$:

$$ 0 \rightarrow \Gamma \hookrightarrow G \twoheadrightarrow Q \rightarrow 0.$$

This problem is concerned with cohomology theory of groups (see the corresponding wikipedia article and probably many books with subject "groups")

No, there are plenty of examples of groups with isomorphic normal subgroups and quotients. Probably, the smallest example: $C_2^2 / C_2 \cong C_2$ and $C_4 / C_2 \cong C_2$.

In general, determining all groups $G$ such that $G/N \cong H$ is quite difficult; this is known as the group extension problem.