Why does the ISDN S0-Bus require terminating resistors?
Simply put:  The old rule of thumb of "you don't need termination if the wire length is less than xx% of the wavelength" is not at all a good rule.  There are many cases where it doesn't apply and ISDN is certainly one of them.  

The rule is wrong, not ISDN.

Another possibility (which does not rule out what I said above) is that some signaling standards operate on what we call a "current loop".  Basically it's the change in current, not the change in voltage, which encodes the data.  In these signaling standards the termination resistor is what provides a path for the current.  With no current, no data is transmitted.

Update:

Let me elaborate on why I think that old rule is wrong...


People misinterpret the rule.  They think that "frequency" means the clock or data rate while it really means the highest frequency in the spectrum of the signal.  Other people will often say things like "it's not the data/clock rate that matters but the edge rate", which is technically correct but doesn't really help either.  The pedantic among us will point out that this is a problem with the person, not the rule-- so sue me.
The rule assumes that you know the frequency content of your signal.  For digital signals with reasonable edge rates you don't really know this.  Further, since good square waves contain an infinite number of harmonics you could interpret this to mean that you always need termination.  Of course putting termination on every signal is silly.  The point is, for digital signals the rule doesn't give you any useful guidance on if termination is needed or not.
The rule assumes that this one signal is the only important signal.  It is possible that crosstalk on this signal will couple noise into another signal, and that adding termination (regardless of trace length) will minimize that crosstalk.


There might be other reasons why this rule of thumb doesn't work, but that's all I have right now.