First, it's absolutely expected that the output current and input current (averaged over a switching cycle) are not equal in a switching converter. If the currents were equal, the efficiency couldn't be any better than a linear regulator's.
Now, let's look at a simple buck regulator:
When the switch (Q1) is closed the input current does indeed go into the load. But part of it also goes to recharging Cout, whose voltage has drooped during the "off" part of the cycle.
When the switch is open, the load still receives current, but it's supplied by D1 and Cout.
So there's no concern that by not drawing input current during part of the cycle you might fail to provide power to the load. It's just part of how a buck converter works.
Would a large capacitor on the input of my step-down converter do the trick?
A large capacitor (Cin in the schematic) won't change the fact that when the switch is open, no input current is drawn.
What it will do is, when the switch is closed, allow much of the input current to come from Cin instead of from the upstream voltage source. This current will be flowing in a relatively small loop, and so not produce so much EMI issues as if it had to flow from the upstream source, however far away that might be.
It also means that whatever inductance there is in the lines from the upstream source to your circuit won't cause Vin to droop during the "on" part of the switching cycle and interfere with the converter's operation.
Edit
I realized you're worried about the peak current drawn during the "on" part of the cycle being higher than the PoE can supply.
Yes, a larger Cin will help with that, by smoothing out the current drawn over the switching cycle. But basically any capacitance on the load side of the PoE will also help.
The choice of operating frequency and L1 value will also affect the peak current draw on the input.
Be aware of the capacitance to ground on each pin of the digital pot. The spec you embedded in your question suggest a range in the order of about 100pF. I don't know the exact switching frequency of your proposed application but, if it were 200kHz, 100pF would have an impedance of about 8 kohms (reactive) and this might be significant compared to the feedback resistor values you intend to enact with the digital pot. This could mean that you get a highly unstable circuit (worst case scenario) or, a lot of ripple voltage (more probable scenario) or absolutely no problem at all.
There isn't enough information in the LT data sheet to call this one so I'd advise setting it up with standard resistor values to cover the range you expect and testing it by adding 100pF caps across those resistors just to see if they introduce oddball behaviour. Other than that it should work just fine.
If it does got belly-up on you there is another alternative and that is to "offset" the feedback voltage with a current sink/source thus "conning" the LT8705 into thinking it needs to raise (or lower) it's mark-space ratio. The current sink/source can be controlled via the pot you want to use and because the pot would be "isolated" from the actual feedback node the capacitance problem wouldn't be an issue. Current sink/sources can be designed with very low capacitances (sub 5pF) so this should be OK. Remember, the standard feedback resistors won't introduce more than 1 or 2 pF and a bit of track might add another 2 pF hence my concern.
Best Answer
This is very nearly the description of the first half of a common PC power supply. You could learn how they work, and then use many of the parts from one. But they are not simple.
What that part of the circuit does is rectify the input to a high voltage DC. The resulting DC is then run through a switching supply with isolating transformer to get the desired voltages. Regulation is done by looking at the output voltage and using it to control the switching system upstream.
It is rather involved, but this may be necessary in order to get the wide input range you seek.