What is needed to specify a group? I have come across several groups, some of which have the same number of generating elements and of the same orders. Take, for instance, D_{2n} and S_n.

Step 1
Note that we can always add more (redundant) generators to the representation without changing the group, the easiest example is ${\mathbb{Z}}_{pq}$ for different primes. It can be defined as:
$${\mathbb{Z}}_{pq}=⟨<!--\; ⟨\; -->a\mid <!--\; \mid \; -->a^{pq}=e⟩<!--\; ⟩\; -->$$
$${\mathbb{Z}}_{pq}=⟨<!--\; ⟨\; -->a,b\mid <!--\; \mid \; -->a^p=b^q=e,ab=ba⟩<!--\; ⟩\; -->$$
I will now explain why this happens:
The problem with groups is that unlike vector spaces, it's not always possible to find a linearly independent set once you have a generating set. We have two important concepts in linear algebra:
1. A generating/spanning set
2. A linearly independent set
The first concept gives you the ability to write every element of your vector space as a linear combination of a few "generators". Notice that a vector space with its vector addition operation forms an Abelian group and therefore, $\stackrel{\to <!--\; \to \; -->}{v}={\mathrm{\alpha <!--\; \alpha \; -->}}_1\stackrel{\to <!--\; \to \; -->}{v_1}+\mathrm{\alpha <!--\; \alpha \; -->}\stackrel{\to <!--\; \to \; -->}{v_2}+\cdots <!--\; \cdots \; -->$ is in fact another way for writing $\stackrel{\to <!--\; \to \; -->}{v}={\stackrel{\to <!--\; \to \; -->}{v_1}}^{{\mathrm{\alpha <!--\; \alpha \; -->}}_1}\cdot <!--\; \cdot \; -->{\stackrel{\to <!--\; \to \; -->}{v_2}}^{{\mathrm{\alpha <!--\; \alpha \; -->}}_2}\cdots <!--\; \cdots \; -->$ in the additive notation instead of the "multiplicative notation". Well, we never use the multiplicative notation in this case but the point is that the concept of a generating set for a vector space is the same as a generating set for a group in essence.
It happens that in linear algebra when we are working over a field, every generating/spanning set contains a linearly independent subset. This gives us a unique representation for every element of the vector space. Failure of this property in an algebraic structure results in the existence of several different representations for the same element, which makes it tricky to define well-defined homomorphisms on it.
But if we indeed have a basis, i.e. a generating set that is linearly independent, we get a well-defined morphism just by determining where each generator should be mapped to. Because of the uniqueness of representation of an element, this gives us a well-defined morphism and we get to completely determine linear transformations (i.e. homomorphisms) easily.
Step 2
What is missed in Group Theory is the concept of a basis. In fact, finite groups cannot have a basis over Z. This leads to the concept of a free group which is way trickier than a vector space. Therefore, when you define G by such a representation, you will not immediately have a way to see what other groups are isomorphic to it. However, I believe you can always add enough relations between generators to uniquely define a group, provided that one exists at all. For example, if $G_1$ and $G_2$ satisfy the same relations, but they are not isomorphic, there must be an equation in $G_1$ that does not hold in $G_2$ and adding that equation to our defining relations should separate these two.

Step 1
Consider two groups $G_1$ and $G_2$, each of which is generated by two elements a and b. In $G_1$, a and b satisfy these relations:
- $a^2=e$;
- $b^3=e$;
- $ab{a}^{-<!--\; -\; -->1}=b^{-<!--\; -\; -->1}$
In $G_2$, a and b satisfy these relations:
- $a^2=e$;
- $b^2=e$
- $(ab{)}^3=e$.
So, the two presentations are quite different. However, $G_1\simeq <!--\; \simeq \; -->G_2$, because, in fact, both of them are isomorphic to $S_3$.
Step 2
Here's another example. This time, I will provide two presentations of the group ${\mathbb{Z}}_6$:
1. group generated by a and the relation $a^6=e$;
2. group generated by a and b and the relations $a^2=e$, $b^3=e$ and $ab=ba$.