Product of conjugacy classes - is there an analog of Tanaka-Krein reconstruction ? Consider a  finite group G. The product of conjugacy classes can be defined in natural way just by multiplying the representatives and counting multiplicities (see e.g. MO 62088). So we get ring with a basis and structure constants are natural numbers. Similar to what one has for product of irreps.
There are many analogies between conjugacy classes and irreps in particular see this article.
Tanaka-Krein duality states that group can be reconstructed from the  tensor category of its representations which is semisimple for finite groups, and hence carries the same information as ring + basis of irreps.
Question:  Can one reconstruct a group having  (ring + basis) made of  conjugacy classes ?
If not -  what partial information (e.g. character table) one can get ? 

Question: Is there any relation between this ring and ring of irreps of the same group ? or may be some other group ?
(Remark. For abelian group they are isomorphic.)
Question: Are there any further analogies between ring of irreps and conjugacy classes except mentioned in the paper cited above ?
 A: 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 less information than 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.
A: The answer to your first question is negative. For a concrete example, you can show that the conjugacy class rings of the nonisomorphic groups $Q_8$ and $D_8$ are isomorphic, via an isomorphism that pairs off the bases as follows: $[1] \leftrightarrow [1]$, $[-1]  \leftrightarrow [r^2]$, $[i] \leftrightarrow [r]$, $[j] \leftrightarrow [s]$ and $[k] \leftrightarrow [rs]$.
As to your question about the relationship between the conjugacy class ring and the character ring, there are lots of partial results that can be stated. Nonetheless, the answer to the question of when these two rings are isomorphic is completely known. This turns out to be the case if and only if the group is $p$-nilpotent with abelian Sylow $p$-subgroup. More generally, for arbitrary finite groups $G$ and $G'$, the following two conditions are equivalent.


*

*The character ring of $G$ is isomorphic to the conjugacy class ring of $G'$.

*$G$ and $G'$ are $p$-nilpotent groups with abelian Sylow $p$-subgroups. Moreover, if $g_1, \dots, g_l$ and $g_1',\ldots, g_{l'}'$ are complete sets of representatives for the conjugacy classes of $p'$-elements of $G$ and $G'$, resp., and if $D_i$ and $D_i'$ are Sylow $p$-subgroups of $C_G(g_i)$ and $C_{G'}(g_i')$, resp., then $l=l'$ and $D_i \cong D_i'$.
This is due to Saksonov, The ring of classes and the ring of characters of a finite group. Mat. Zametki 26 (1979), no. 1, 3–14, 156.
A: As suggested in the comments, it has been known almost since the beginning of the representation theory of finite groups that knowledge of the character table is equivalent to the knowledge of the "class algebra constants", which are the structure constants telling you the multiplicity with which the class sum $C_{z}$ occurs in the product of class sums $C_{x}C_{y}.$ A formula of Burnside tells us that this multiplicity is $\frac{|G|}{|C_{G}(x)||C_{G}(y)|} \sum_{\chi} \frac{\chi(x)\chi(y) \chi (z^{-1})}{\chi(1)}$, where $\chi$ runs through the comples irreducible characters of $G.$ In the other direction, the centre of the complex group algebra
has a basis consisting of primitive idempotents. There are several standard ways to recover this basis given the explicit knowledge of multiplication in the algebra, which is given by the class algebra constants. These primitive idempotents are in bijection with the irreducible characters, and the primitive idempotent $e_{\chi}$ corresponding to the irreducible character $\chi$ can be written as $\frac{\chi(1)}{|G|}\sum_{y} \chi(y^{-1}) C_{y}$, where $y$ runs through a set of representatives for the conjugacy classes of $G$ and $C_{y}$ denotes the class sum of $y.$ It is also worth noting that the class sum $C_{y}$ may be expressed as $\sum_{\chi} \frac{[G:C_{G}(y)]\chi(y)}{\chi(1)} e_{\chi},$ where $\chi$ again runs through the irreducible characters.
A: One nice fact, I think, is that the formula of Burnside that Geoff Robinson gives above,
$$
 N^{C_z}_{C_x C_y} = \frac{|G|}{|C_G(x)| |C_G(y)|} \frac{\chi(x) \chi(y) \chi(z^{-1})}{\chi(1)} 
$$
can be understood nicely from a geometric / topological quantum field theory perspective. I think it is precisely the "Verlinde formula" for the modular category Rep(/\G), the representation category of the Drinfeld double of C[G]. The Verlinde formula says that in a general modular category, we have
$$
 N^k_{ij} = \sum_r \frac{s_{ir} s_{jr} s_{k^* r}}{s_{0 r}}
$$
where $s_{ij}$ is the S-matrix. 
More concretely, this is to say that it has a natural interpretation in terms of G-bundles on the torus. This perspective also comes out in the appendix of Zagier to "Graphs on surfaces and their applications" by Lando and Zvonkin. 
