The theory is good (mostly) the execution is the problem.
If the APDS does not separate cleanly and consistently accuracy turns to crap. It doesn't matter how good the penetration is if the majority of rounds fired do not hit the target.
Agreed, I even explicitly mentioned that in the first sentence of my post: "...had APDS been developed into a more mature state in the run up to WWII...". An APDS design is indeed pointless if it can't hit the proverbial barn from the inside.
We can play with the math a bit more and see where we wind up for Joules per sq cm of plate area to find out what is needed for steel cores to work, assuming there is no problem with the steel cores shattering.
The French were trying to improve some pretty low performance guns, tying to improve even 600-700ms guns gets a lot harder because you are trading away the weight per unit of frontal area for velocity.
Shot size..............weight............................grams per unit of frontal area.
40mm......................1088g.................................86.62
50mm......................2050.................................102.13
57mm......................2720.................................106.66
75mm.....................6800...................................154
A the same velocities it is the weight per unit of frontal area that is important until we get into exact heat treatments/hardness and failures of material.
Let's play at bit more. So assuming we start with a 75mm gun that can shoot a traditional full-bore AP(C(BC)) shell at 600 m/s (keeping nice round numbers to keep the math simpler). Let's use the 6.8kg weight you quote, which should be in the ballpark for a WWII era APCBC shell. Now if we instead want to shoot a subcaliber round at 900m/s, keeping the kinetic energy constant assuming that's roughly proportional to recoil, barrel pressuer etc., we need the weight of the subcaliber round to be 4/9th of the weight of the full bore shell, that is 3.0kg.
To determine the required diameter of the 3.0kg shell, let's assume it's the same shape and density as the full bore shell. This gives us the diameter of the subcaliber shell as the diameter of the full-caliber shell times the cube root of the relative masses. Thus 57mm.
Note there is some inaccuracy here. The recoil is not exactly proportional to the kinetic energy. And the mass of the sabots has not been taken into account. And further the APDS shot is denser as it doesn't have a cavity filled with low density HE. And probably the shape of the APDS projectile can be optimized somewhat as it doesn't have to take into account need for driving bands etc. But let's assume we're in the ballpark. That is, a 6.8 kg 75mm shell at 600 m/s vs. a 3.0kg 57mm shell at 900 m/s.
Now to determine the penetration. If we use the Thompson F-formula, if we rearrange it a little bit we can see that the penetration T is proportional to the kinetic energy divided by the frontal area, times some unit conversion factors, obliquity factors and some semi-empirical correction factors (the "F-factor" which includes shell shape, material etc. etc.). So in both cases we have the same kinetic energy (at the muzzle), but the subcaliber shell has much less frontal area. And thus the penetration will be higher, by a factor of (75/57)^2 = 1.7.
Now if we switch to tungsten, the big advantage here is that we can get the same KE in a narrower shell, thus increasing the penetration. Let's use tungsten semicarbide, which AFAIU was used during WWII, with a density of 17.15g/cm^3, vs. 7.85g/cm^3 for steel. Doing the same calculation as above, but taking into account the difference in density, we get d2 = d1 * cbrt(m2/m1 * rho1/rho2) = 44mm. Plugging that into the penetration calculation, we get compared to the full-bore shell a penetration a factor of (75/44)^2 = 2.9. So quite a big difference indeed. And of course, tungsten carbide doesn't have the same problem of shattering as steel, so no need to limit ourselves to a MV of 900m/s. We could increase the MV and reduce the diameter and get even better results. Left as an exercise for the reader.
And further, as you say, the narrower tungsten penetrator will also lose velocity slower. Assuming the rate of velocity loss is proportional to the KE/frontal area similar to penetration, (57/44)^2 = 1.7. So the tungsten penetrator can fly a distance of 1.7 times further than the steel penetrator for the same loss of velocity.
So yes, clearly tungsten is superior. But per the above, steel APDS is still better than traditional full-bore APCBC.