A Micro Four Thirds sensor is one fourth the size of a full frame sensor. To view both captured images at the same display size, the image from the µ4/3 camera must be enlarged by a factor of four compared to the enlargement of the image from the FF camera. But the light captured by the smaller sensor is not striking the sensor at 1/4 the wavelengths and at 4X the density of the light striking the FF sensor. Even if it were, the properties of silicon would still be the same and the CMOS chip would not be capable of absorbing 4X as many photons per unit area before reaching full well capacity.
My understanding is that low-light performance is a function of the amount of light received on the sensor per unit area (correct if wrong).
You're not entirely wrong, but only partially correct. Low light performance is a function of electrical signal produced by the amount of light received on the sensor compared to the 'noise' (both electrical signal produced by anything other than light falling on the sensor AND variation in the exact amount of light falling on adjacent discrete units¹ of the sensor due to the random nature of light). In other words, low light performance is based on the ratio of signal to noise. If there were no noise included in the electrical signal from a sensor, we could amplify a weak signal (i.e. a signal produced by 'low light') as much as we wanted with no degradation in quality.
¹Each discrete unit is called a 'sensel' or 'pixel well'.
The capacity of a silicon semiconductor, such as a CMOS imaging sensor, is based upon being able to measure a specific number of photons striking it per unit area. If you make a sensor that is 1/4 the size but has the same resolution then each sensel ('pixel well') is also 1/4 the size of the sensels of the larger sensor. With 1/4 the area, the silicon can only collect 1/4 as much light before it reaches 'full well capacity'. Once a sensel had reached full well capacity any additional light falling on it can not be measured, and thus has no effect on the resulting readout for that sensel (the excess voltage may jump to adjoining sensels in a process we call 'blooming', but no additional usable information will be collected - in fact less will since the blooming will mask the amount of light that was actually measured by the adjacent pixels.)
This is why a larger sensor with larger sensels will have higher dynamic range, all else being equal. Each sensel can absorb more photons before reaching full well capacity.
The smaller size of each sensel generally has the effect of increasing both types of noise described above.
"Read noise" is that noise created by electrical components of the sensor and any other part of the camera's circuitry used to transmit the signal before it is converted to digital information. For a sensor of a given generation of technology, the amount of read noise generated per pixel is the same regardless of the size of the pixel well. Since the maximum signal of a smaller sensel is less than that of a larger sensel, the ratio of read noise to signal will be greater with smaller sensels.
"Poisson Distribution noise" or "Photon Shot noise" is the noise created by the random distribution of photons in a light field. Because photons travel in a wave form rather than a straight line, as an equally bright area of light strikes an imaging sensor the numbers of photons collected by each sensel will not be identical. Rather than a uniform pattern of photons striking the sensor, the pattern will look more like a pattern of birdshot fired from a shotgun. The smaller each sensel is, the more the effect from this random distribution of photons will be. The larger each sensel is, the less the effect will be as the differences are averaged out by the larger sample size of each sensel. Shot noise is further complicated by the way that Bayer masked sensors interpolate color values from sensels that only measure the total number of photons that strike each one, regardless of the wavelength of each photon.
For example, if using a 50mm/1.4 lens on a FF and a 25mm/1.4 lens on the MFT, the FF will gather four times as much light, but it will spread that light over a sensor that is four times as big, so won't the amount of light per unit area on the sensor remain the same?
The amount of light (and thus signal) per unit area would remain the same. But the amount of noise would be higher assuming equal resolutions.
Remember, the other consideration to be made is that the Micro Four Thirds sensor is one fourth the size of the full frame sensor. To view both captured images at the same display size, the image from the µ4/3 camera must be enlarged by a factor of four as well. Enlarging the image also increases the perceived effect of the noise.