You can construct the ring of conjugacy classes from the character table in the following manner:

First note that the ring of conjugacy classes tensored with $\mathbb C$ can be identified with assignments of a number to each irrep, or column vectors, with entrywise addition and multiplication. So we have to find a basis. For each column vector of the form: 


$\left(\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{array}\right)$

we know how to write it in terms of conjugacy classes: as the corresponding row in the character table, divided by the order of the group. 

So we invert that matrix to find out how to write the conjugacy classes in terms of irreps.

Concretely, the ring of conjugacy classes is the ring generated by the columns of the inverse transpose of the character table times the order of the group under entrywise multiplication.

Thus, it gives no more information than the character table.

Edit: It also gives no information then the character table. The reason is that one can reconstruct the center of the group algebra from this ring by tensoring with $\mathbb C$. One then has one minimal idempotent for each irreducible representation. We can find the minimal idempotents algebraically and compute how to write them in terms of the conjugacy classes, which tells us the character table.