My Buck-Boost Converter Circuit Isn't Working
Your DC-DC won't work at all without at least one of the feedback transistors being turned on.
When they are turned on the VCE(sat) is directly in series with your feedback voltage. While VCE(sat) will likely only be a few mV in this case it still has an impact on the accuracy of you V(out) 

You should be extra careful that the transistors used in the feedback loop are extremely low leakage and not too high an Hfe. You don't seem to specificity the part number in your schematic so I can't check. 

The CE leakage current when these transistors are off (no Base current) is typically higher than that quoted for the CBO leakage specified in the datasheet. The feedback resistors indicate that the feedback current is very small at only about 5uA. With only one of the three feedback switches on, it means you are dealing with 2x the leakage current.

If you are NOT driving the transistor bases, then the CE leakage will be approximately Hfe*I(CBO). With high Hfe devices this can be troublesome. 
You should ensure that the drive point for your base resistors is always driven by a logic zero or one, this ensures that the base is terminated to close to ground when turned off.

You might find this answer helpful. 

Update: Now you have added the feedback switching transistor data it is clear that they are unsuitable for this task. They are power devices with a max I(CBO) of 1uA so even when off will impact the feedback loop. 

If you can find something in the same package layout, then you are looking for characteristics similar to these ONSemi small signal devices, NST3904DP6T5G.   

 

With I(CEX) of only 50nA these are a much better choice, and you can find transistors with I(CEX) down to 10nA. I(CEX) is the CE current with the base open circuit so is a much better indicator of leakage performance than I(CBO).

Ideally IMO the best choice choice for a switched feedback loop would be a small signal Depletion mode N-Channel FET switching from the output.