You could just go old school with an AC linear power supply
They make small transformers that you can mount on a prototyping board .. add a fuse, a bridge rectifier,a smoothing capacitor and a linear regulator to regulate to 5v and a small capacitor on output of this regulator (or capacitor - inductor - capacitor as a pi filter) and you have a linear power supply.
Pick a transformer that outputs 8..12v AC because after rectification using a bridge rectifier, you'll have a DC voltage with a peak equal to Vdc peak = Vac x 1.414 - 2 x (voltage drop on diode inside bridge rectifier)
So for example, for a 9v AC output transformer, you'll have Vdc peak = 9vx1.414 - 2 x 0.8v = 11v
If you use a plain linear regulator, such regulators usually require that the input voltage must always be above the output voltage by at least 1 - 1.5v so let's make sure it will by adding a capacitor after the bridge rectifier.
You say you need around 0.1A so let's calculate things for a maximum of 0.25A of current.
This formula approximates how much capacitance would be the minimum to use to always have the voltage above some threshold :
C = Current / [ 2 x AC Frequency x (Vdc peak - Vdc min) ]
With the 9v transformer example above we can put the values in formula C = 0.25A / 2 x 60 Hz x (11v - 7v) = 0.25 / 120*4 = 5.20e-4 Farads or around 520uF so to guarantee at least 7v before the regulator even if your circuit consumes 0.25A, you would need at least 560uF.
Considering you're estimating 0.1A, you can probably go with a standard value of 470uF (should be rated for at least 25v).
For the actual linear regulator, it won't really matter ... a 7805 would work... and have a small 100uF 10v (or higher) rated capacitor on the output of this regulator.