Question about discrete log - why can one attacker not solve for g but two colluders can?
The idea is that someone who just knows $g^d \bmod N$ and $d$, they cannot compute $g$ (because they don't know the factorization of $N$; hence this is the RSA problem); however, if they know $g^d \bmod N$ and $d$ and $g^e \bmod N$ and $e$, then (assuming $gcd(d, e) = 1$), they can (and once they have $g$, they can compute $g^{abc} \bmod N$).

That's because, since they know $d$ and $e$, they can compute two integers $s$ and $t$ such that $sd + te = 1$ (that's where the assumption that $gcd(d, e) = 1$ comes in; if that's not true, then no such $s$, $t$ will exist).  Once that have those two integers, they can compute $(g^d)^s \times (g^e)^t = g^{sd + te} = g^1 = g$, recovering $g$.  One of $s$ and $t$ will be negative; that's not an issue, as we can compute inverses modulo a number of unknown factorization.