Edit: Unfortunately, the first version of this answer was wrong, so I had to rewrite it almost entirely.
So the formula you want to prove is:
$$
\sum_{n=1}^\infty C_{n,q} t^n =\prod_{k=1}^\infty \frac{1-t^k}{1-qt^k}
$$
where $C_{n,q}$ denotes the number of conjugacy classes in $\textrm{GL}_n(q)$.
Note that
$$
\frac{1-t^k}{1-qt^k} =1+\sum_{i=1}^\infty (q^i-q^{i-1}) t^{ki}
$$
As Paul Garrett notes in his answer, $q^i-q^{i-1}$ is the number of monic polynomials in $\mathbb F_q[x]$ of degree $i$ with non-vanishing constant term. One usual way of counting conjugagy classes of matrices in $\textrm{GL}_n(q)$ (or $\textrm{Mat}_n(\mathbb F_q)$, if one admits the eigenvalue $0$) is counting isomorphism classes of $\mathbb F_q[x]$-modules of dimension $n$, which by the structure theorem on f.g. modules over PID's are given by
$$
\mathbb F_{q}[x]/(p_k(x)) \oplus \mathbb F_{q}[x]/(p_k(x)\cdot p_{k-1}(x)) \ldots \oplus F_{q}[x]/(p_k(x)\cdot p_{k-1}(x)\cdots p_{1}(x))
$$
where $p_1,\ldots,p_k$ are monic polynomials with constant term $\neq 0$ (uniquely determined by the isomorphism type of the module) such that $1\cdot \textrm{deg}(p_1)+2\cdot \textrm{deg}(p_2) \ \ldots + k \cdot \textrm{deg}(p_k) = n \textrm{ (the dimension of the module)}$. Note that some of the $p_i$ may very well be equal to $1$.
From the above considerations it is clear that we can parametrize isomorphism classes of $\mathbb F_q[x]$-modules by giving a sequence of monic polynomials which eventually becomes constant $=1$, by letting the module given above correspond to the sequence
$$
(f_1,f_2,f_3,\ldots) =(p_k, p_{p-1},\ldots,p_1,1,1,\ldots)
$$
Note that in this sequence, $f_m$ constributes $m\cdot \textrm{deg}(f_m)$ to the dimension of the module. It follows
$$
\begin{array}{rcl}
\sum_{n=1}^\infty C_{n,q} t^n &=& \prod_{m=1}^{\infty}\left(\sum_{i=0}^\infty \#\textrm{(choices for $f_m$ that contribute $mi$ to the dimension)}\cdot t^{mi}\right)\newline &=&\prod_{m=1}^{\infty} \left(1+\sum_{i=1}^\infty (q^i-q^{i-1}) t^{mi}\right) =\prod_{m=1}^\infty \frac{1-t^m}{1-qt^m}
\end{array}
$$
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.
Best Answer
If $c(G)> 5|G|/8$, then the average character has a dimension-squared of less than $8/5$, so at least $4/5$ of the characters are dimension $1$ (since the next-smallest dimension-squared is $4$), so the abelianization, which has one element for each 1-dimensional character, is more than half the size of the group, so the commutator subgroup has size smaller than $2$ and so is trivial.