I design antennas and wireless links for a living, sir. Yes, I have done this. Since 1984.
Of course 900 MHz has better range, if one is using dipole antennas. The point is, the apparent difference in range is due to the antennas used, not some intrisic property of the propagation medium. (It's also likely that the 5 GHz transmitter has lower output power, and the 5 GHz receiver a higher noise figure, than the 900 MHz versions, making the 5 GHz range even less, but I'll ignore those factors for now.)
Give me parabolic dishes, and 5 GHz will go where 900 MHz will not. Give me resonant dipoles, and 900 MHz will go where 5 GHz will not. The behavior follows the antenna selection. It has nothing to do with "propagation," whether through trees, concrete block, or anything else.
Maybe numbers will help. Since the effective area goes as the wavelength squared, the amount of power captured from a resonant 5.5 GHz dipole, compared to that from a resonant 900 MHz dipole, from a given power flux density, is (900/5500)^2 = 0.027, or 2.7 percent. This means that, with identical efficiencies, matching loss, etc., the signal the 5.5 GHz receiver will get from its antenna is down 10*log10(0.027) = 15.7 dB from the signal the 900 MHz receiver gets. That's why your range at 5 GHz is less than at 900 MHz.
Now, let's move to parabolic dish antennas. Since the effective area of the dish antenna is the same for both bands, for a given power flux density the same signal level will be presented to each receiver at the antenna terminals. However, the power flux density generated by the transmit parabolic dish antennas will vary, but this time by the frequency squared, so the power flux density at the receivers on 5500 MHz will be 15.7 dB stronger than it will be at 900 MHz.
You can see where this is going. One has four possibilities:
1. Tx dipole, Rx dipole: 5.5 GHz signal 15.7 dB weaker than 900 MHz
2. Tx dipole, Rx dish: 5.5 GHz signal the same as 900 MHz
3. Tx dish, Rx dipole: 5.5 GHz signal the same as 900 MHz
4. Tx dish, Rx dish: 5.5 GHz signal 15.7 dB stronger than 900 MHz
You may ask how much range change is implied by a 15.7 dB change in signal strength. One simple range model is the use of a loss exponent; a typical value for the loss exponent might be 3 or 4. If we are optimistic, and take 3 as the value for the loss exponent, and assume a 15.7 dB change in loss, one can determine the ratio of the two distances as
10 * n * log10(d1) - 10 * n * log10(d2) = loss(dB)
30 * log10(d1/d2) = 15.7
d1/d2 = 10^0.524 = 3.34.
The band with the 15.7 dB strength advantage (the 900 MHz band, when dipoles are used, or the 5 GHz band, when dishes are used) has more than three times the range of the alternative. (If one assumes a loss exponent of 4, as might happen in an ugly indoor environment, the ratio works out to be about 2.47.)