WARNING LONG POST AHEAD!!!
IDK about back then when things weren't digitally stored, but in the digital age, it may be better to record at higher volumes rather than low. Let's go back and take a look at how a recording is done digitally today, going to the core basics. A little above 0s and 1s though. Given a constant time, we will measure the amplitude of a given wave at a given constant time. Essentially, we create a drawing of a wave this way. This wave can then be defined by some sort of function f(t) such that for a given time, the wave is at x frequency. For simplicity sake, let's assume we are always in positive integer system + 0.
Now, we can always reduce the wave with minimal loss of the actual structure of the wave itself. Let's say that f(20) = 25.4433. Now let's say that the recording company was using something that halved the volume (recorded at half volume). We'd get the reading f(20) = 12.72165 = 12.7266 (4 digit arithmetic). Note how we lost some precision in the lower volume. Now the error in both instances will always vary by +/- .00005. This we'll call absolute error. However, relative error is the more important factor here. Relative error = | actual - calculated | / actual where actual is the actual number it should be and calculated is what would be calculated. | actual - calculated | = absolute error. We know the absolute error for both instances, .00005 (differs by computer though). So we get | .00005 | = | -.00005 | = .00005 for absolute error. This means that the absolute error is the same in both instances. We can calculate the relative error now. We get the following relative errors: .00005 / 12.72165 and .00005 / 25.4433 or .000003930 and .000001965 respectively. Notice how the error in the smaller amplitude is twice that of the one recorded at the higher amplitude (that is twice as high in amplitude). Digitally speaking, it is more adventageous to record at higher SPL so long as we don't induce distortion or clipping.
I hope all of that made sense. We can generalize it like this. Assume we have two SPL levels: spl1 and spl2. Assume they aren't the same (one is higher than the other). spl1 > spl2. The absolute error for both is based upon machine precision (computers percision it can calculate to). It would be +/- u where u is the machine's precision. u / spl1 < u spl2 iff spl1 > spl2. The relative error of the higher volume will always be better than the lower volume mathematically speaking.