I've seen way too many implementations where they simply send all the current from the local PS to the device and mark off whether that particular node is powered or not (and sometimes, incorrectly).
If you design the PWM PSU you should choose very high switching speeds and it will make noise production a non-issue. After all, USB power is the output of yet another PWM supply, not like that's going away...
With a linear solution, regardless if you use a chip or zener solution, you'll get less than 33% efficiency. But if you only need 100mA (for bus-power) or 500ma (local PSU) then it's probably good enough. Sounds like you need around 200mA or so to deal with surges thus you will need the local PSU and bus power is insufficient.
So 5V - 1.5V = 3.5V drop by resistor.
3.5V=200mA * R
R=17.5 ohms. Round to 15 ohms. P=v^2/r=3.5*3.5/15=.8 watts wasted here, round up to requiring a 1W device.
Zener diode dissipation worst case: recalculate current with rounded resistor:
3.5V=I * 15 = 233mA
power dissipated = V * I = 1.5V * 233mA=350mW, make sure you get a 500mW zener diode. Looks like it is feasible to get something like this working with available parts.
(If you needed 1A of 1.5V, you'll have to use a switch mode PSU. With a properly designed SMPSU you can get 1300mA or more at 1.5V without too much difficulty from a local powered USB port. (around 250mA with bus-powered) This sounds like overkill though based on your observations, unless you want to be able to run off of bus power.)