Thinking about this again, the current through the resistor and therefore the voltage should be kept as low as possible, to prevent the power dissipated in the resistor heating it up and interfering with the temperature measurement. In any case, once the digital potentiometer is in series with 1k1, it won't see the full voltage, so it won't be an issue.
If it measures the resistance using a constant current source, then the voltage will be proportional to the resistance, so simply connecting it to a voltage source of the correct value should be able to fool it.
If it works by using a voltage source, then it can be fooled with a current source.
If it works using a potential divider, then it's more tricky. Yes four 1k digital potentiometers can be connect in parallel to make 250R.
I've had another idea of using an op-amp to effectively scale the value of the digital potentiometer down to a lower value. 1024 position 20k, 1% tolerance, potentiometers seem to be widely available in Digikey. Although it's more components, it might be easier, as there's only one digital potentiometer to worry about, which will also carry a fraction of the current. It will probably require a frequency compensation network to avoid oscillation. I haven't done any dynamic analysis on, whichwill depend on the op-amp and MOSFET used.
The horizontal axis is the digital pot's resistance and the vertical R_OUT, which is the output. It's measured using a constant current source and applying Oham's law. I1 represents the current source in the device, but it could be a voltage source and it will still work.