Interesting proposal. My main concern is that the composition of the remaining extract will be vastly different between a 1.048 beer brewed with S. cerevesiae vs the S. ludwigii. With so much maltose remaining, I'd be concerned that it could taste sweet, underattenuated and/or "worty". Certainly worth a try, though.
NA beer is sweet because it is targeted at Standard American Lager drinkers, which is a consumer base that is generally not into hoppy beers. I think that the residual sweetness could be balanced via the addition of bitterness. While maltose is sweet, it is not glucose or sucrose sweet.
Maltose Disaccharide 0.33 – 0.45
Glucose Monosaccharide 0.74 – 0.8
Sucrose Disaccharide 1.00 (reference)
Fructose Monosaccharide 1.17 – 1.75
I believe that the approach I outlined above would work better with a grainy tasting malt like Pilsner malt than it would with American 2-row. If we are simulating a green bottle beer, then apparent extract could be reduced. Green bottle beers are fairly well attenuated. Let's say that the OE is 11.5P (O.G. 1.046) and the AE is 2P (F.G. 1.008)
RE = 0.1808 x 11.5 + 0.8192 x 2 = 3.72P
As calculated above, RE is going to be approximately 92% of the OE.
OE = 3.72 / .92 = 4.04P = ~1.016 S.G.
A 4P wort contains 40 grams of extract per liter, which means that one liter of low-alcohol beer produced from a 158F mash would contain 40 x .41 = 16.4 grams of maltose. Since maltose is 1/3rd as sweet as sucrose, 16.4 grams of maltose is roughly equal to 16.4 x .33 = 5.4 grams of sucrose in sweetness, which is equivalent in sweetness to about 1.4 teaspoons of sucrose dissolved into 1L of water.
Maltotriose is technically the shortest saccharide polymer that can be classified as a maltodextrin. Have you ever tasted maltodextrin powder? It is not very sweet.