I know that a digital signal with sharp edges (going from one y-Value to another at the same x-Value) will have infinite bandwidth.
A digital signal never has "sharp edges"; by the virtue of being digital, it's only defined at discrete points in time. So, it's just a sequence of numbers; these numbers don't have "edges" in between; the space in between two samples is simply undefined, has no value.
It does not have infinite bandwidth, by no understanding of the word I'd subscribe to. Being digital, the "whole world of bandwidths it could have" is the Nyquist rate. So, no.
Only when we take a signal composed of a comb of dirac deltas in the time-continuous domain we see a periodic spectrum – indeed, infinite bandwidth. But that's not the original digital signal, that's a very specific conversion of it. For time-continuous signals, we actually have infinitely extending spectrum.
Usually, and more usefully, we assume that, if there is a time-continuous signal equivalent to the digital signal (that's not generally the case! It helps trying to remind oneself of that, once in a while), that it was bandwidth-limited to begin with – otherwise, no digital signal could represent it, anyways.
So, I think your premise is a bit confused here.
I read that aperiodic time limited signals will have an infinite bandwidth
In continuous-time, under the continuous Fourier transform,
reconstructed_signal = ifft(fft(some_signal))
You're still in discrete-time domain. So, the statement above has little to do with your statement here!
What is the bandwidth of a random signal?
In continous-time, it could be up to infinite.
In discrete-time, it could be up to Nyquist.
"up to", because "random" doesn't tell us much – you could have a random signal that is weak-sense stationary, correlated and has a limited spectrum in either case.