The basic claim is that if you replaced the total USA48 power generated in 2019 by a scaled up version of the 10% of that which was wind and solar (W&S)), then you would need 250 TWh of storage to make it work, and this is impossible.
It is of course wrong. To explain qualitatively, an issue with both fossil fuel (FF) generation and W&S is that demand varies throughout the year. It is possible to generate just the average required, and use storage to meet the peak. But traditionally, this is of course never done. Enough generation is always provided to meet the peak, so that some is idled at other times. And you would do the same with W&S, with a subtle difference that there is no need to idle outside the peak, since no fuel is required.
What these reports do is to wrongly underestimate the amount of W&S required, so it is not meeting the peak, and so is using storage to cover the annual variation. That accounts for the huge storage estimates. The reason is that they scale up W&S so the average matches the FF average, although I think even then they underscale. But it is the wrong thing to do, because the FF profile was sculpted to match the peaks, by idling at other times. The 10% W&S profile did not have that requirement, and so if you scale it by average, it won't match the peaks.
Quantitative scaling
Ken Gregory has an extensive spreadsheet here which has the basic data I used. It has hourly generation figures of each source for 2019, and also other years, but I'll stick to 2019. Ken calculated a "target" T, which was total generation excluding nuclear and hydro. I'm not sure why the exclusion, but I'll do that too (it isn't much). Then he looked at the difference between this T, and W&S multiplied by a scale factor. I will call the factor H, and the product HW. He used values about H=7. He accumulated the difference HWT, which became the storage S. He showed that if S has to be positive through the year, then it could rise as high as 250 TWh.I use a slightly different, and more realistic approach to storage. I specify a maximum storage Sm. Then at each hourly step, the difference HWT (which might be negative) is added to S only to the extent that Sm is not exceeded. In fact, I set Sm=0, because you can add a constant without changing anything. Then the storage required is the minimum (negative) value reached during the year. Of course, this is the bare minimum for just getting through 2019; a reserve will have to be added to H to allow for less favorable years.
So I did that using various factors H. In fact H=10 is what would scale up W&S to match total demand for 2019. I got the following values for minimum storage required:

It is close to exponential decrease. And in my reckoning, H=15 is closest to matching the existing FF build, and 2.4 TWh is not an impossible amount of storage. But building a bit more W&S reduces this a lot further.
Here is a graph of the various cumulative storages. The xaxis is in hours of the year 2019. You can see that at H=7.3 the storage does have to make up for a big change in annual demand, while at H=10, is is only needed to cover the short term changes, since there is enough generation to cover the peak.
I'll show the same on a log scale, same colors. It distorts the annual cycle, but gives a better picture of the higher H storage behaviour.
Conclusion
Ken Gregory, amplified by David Wojick, claim that a simulation of 2019 electricity generation for USA48 with wind and solar only shows a requirement for very large storage (250 TWh). But as siimilarly simulated here, that is because they provided too little W&S, thus requiring storage to cover the annual demand cycle. Doubling the provision reduces storage to 2.5 TWh, with further exponential decrease.The R code and data used are in a zipfile here
Update  I've added an XL file and a readme.txt to the zipfile.