In principle nothing should speak against this. The SCOUT uses an I.F. of 6.144, the L.O. is generated by a 2.2MHz PTO signal, mixed to and by a NE612 XO/mixer in the band modules.
For 80m that means: 7.444 (XO) + 2.2 (PTO) - 6.144 (I.F.) = 3.500 (QRG)
This would mean for 500kHz: 4.444 (XO) + 2.2 (PTO) - 6.144 (I.F.) = 0.500 (QRG)
It would be necessary not only to replace the crystal but also the LO filter network.
The 160m module can be tuned to about 1.5MHz, however, the PTO stabilizing aka FLS (frequency lock system) does not lock in anymore, Ten-Tec mention here that the AM band rejection in the 160m module would be causing this.
Since the modules also carry the band filter, it seems that no modification to the radio itself is required.
Now we've seen that 4.444MHz is the highest usable local oscillator frequency. So, what crystal will be available and where would the lower band edge will be?
- 4.433MHz => 489kHz
- 4.194MHz => 250kHz
- 4.096MHz => 152kHz
- 4.000MHz => 56kHz