This is not an answer, only a simplification and conjecture concerning that simplification. First, only consider $N$ large enough so that the interval/circle is fully covered (the "eventually" in my comment). Second, rather than your exponential distribution, fix all rods to the same (small) length $L$, perhaps $L < \frac{1}{2}$ suffices. Retain your assumption that the left endpoint of each rod is chosen uniformly in $[0,1]$. Then I conjecture that the limiting distribution is uniform with mean rod length $L/2$. I have only heuristic arguments for this (shorter rods get absorbed by newly added ones, existing longer rods get chopped from the ends). Perhaps you could alter your simulation to this simplified circumstance to see if this is empirically true? (Addendum: I verified this myself, and indeed it seems to hold empirically.)
If this conjecture holds, then perhaps it holds even for an exponential distribution, with $L$ now the mean length of that distribution.
Addendum: I verified this
myself, and indeed it seems to hold empirically. Here are results of a simulation adding
10 million rods of length $L=\frac{1}{10}$ to $[0,1]$. Only lengths of rods within
$[L,1-L]$ are averaged in the graph (to exclude edge effects).
