Enumerating all finite groups is a formidable task and as far as I know an unsolved problem in its full generality. I have thought about what if one approaches this by constructing the groups stepwise by constructing character tables by systematically combining sets of "valid" irreducible representations. That leads me to two basic questions:
Is every group uniquely characterised (up to isomorphism) by the full set of its irreducible representations?
Is every set of "characters" grouped into irreducible representations which fulfills the Schur theorem automatically a "character table" belonging to a group?
Edit: In case the answer to 1. is no: Which information is missing?
I think the answer to Question 1 is yes. The irreducible representations of $G$ over the field $K$ are homomorphisms $\rho_i:G \to M_i$ for $1 \le i \le k$, where $M_i = {\rm GL}(n_i,K)$ for some $n_i$. So we can define a homomorphism $\rho:G \to M_1 \times \cdots \times M_k$ by $\rho(g) = (\rho_1(g),\ldots,\rho_k(g))$.
Provided that the characteristic of $K$ does not divide $|G|$, the intersection of the kernels of the $\rho_i$ is trivial, and so $\ker \rho = 1$ and $G \cong {\rm im} \rho$. I am interpreting your assumption as meaning that $\rho_i(g)$ is known for each $i$ and $g$, and hence so is $\rho(g)$.