Is it possible to prove that a group is the only group of its order given some properties?

Let $G$ be a group of order $6$

Then there are elements of order $3$ and $2$, say $a$ and $b$ repsectively by Cauchy's Theorem.

Let $H=\langle a \rangle $ and $K=\langle b \rangle$

Then $H \cap K=\{e\}$ since any order of any element in the intersection divide both $2$ and $3$.

Now $H$ is normal in $G$ since $[G:H]=2$

Thus $HK $ forms a group and infact $G=HK$

Thus every element in $G$ can be listed as $\{1, a,a^2,b,ab,a^2b\}$

Let us think about the element $bab^-(=bab)$ .

We should note $o(bab)=o(a)$. Which one of the above elements can be equal to $bab$ and what can we conclude from there ?

In general, for given $n \in \mathbb{N}$, there is not some simple way to figure out all non-isomoprhic groups of order $n$ (though it's been deduced for special cases of $n$ and a number of 'small' $n$). See more in this post.

To answer your last question, 'up to isomorphism' means that the objects in question aren't really different in terms of structure, just possibly 'name'. To be more specific, let's say you have two groups $G$ and $H$ as described:

$G$ is the group with set $\{ (0,0), (1,0), (0,1), (1,1)\}$, and with operation $+$ defined by coordinate-wise addition, modulo $2$. So for example $(1,0) + (1,0) = (0,0)$. Notice this group has order $4$, and each non-identity element has order $2$.

$H$ is the group with set $\{1,a,b,c\}$ and operation $\cdot$, defined by the following multiplication table ($1$ is identity of course): \begin{array}{|c|c|c|c|} \hline 1 & a & b & c \\ \hline a & 1 & c & b \\ \hline b & c & 1 & a \\ \hline c & b & a & 1 \\ \hline \end{array} Note that $H$ also has order $4$ with each non-identity element having order $2$.

Evidently, $G$ and $H$ have the same 'structure'. To make this more precise, we can actually construct an explicit isomorphism $\phi: G \to H$. In this case, notice that $G = \langle (1,0), (0,1) \rangle$ and $H = \langle a,b\rangle$. Then define $\phi: G\to H$ by: $$ (1,0) \mapsto a, \; \; \; (0,1) \mapsto b. $$ I'll leave it to you to check $\phi$ is a homomorphism and that it's bijective (should be very straightforward from definitions).

Since $\phi$ is an isormorphism between $G$ and $H$, we say $G$ is isomorphic to $H$, and in an algebraic sense there is really no difference between $G$ and $H$, just the 'name' we gave the elements. But the relationship between the elements is identical. Both of these groups are really just $\mathbb{Z}_2 \times \mathbb{Z}_2$.