It is, indeed, not differentiation in the sense you're used to. Here it just means a map that satisfies the Leibniz rule, (fg)' = f' g + f g' or D(fg) = D(f) g + f D(g). Maps with this property are usually called "derivations".
Although you might also want to consider that "integration" (really, indefinite integrals, AKA antiderivative) is only defined up to a constant. So why couldn't it be the same for this "number derivative"? Perhaps the "antiderivative" is only defined up to something. It'd be a fun exercise, if you're interested. Can you figure out under what conditions do you get D(a) = D(b)? Put differently, given an integer c, what are the solutions to D(x) = c?
Solutions to D(x)=C is probably something like p_a^a * p_b^b * … * p_n^n with a+b+d+…+n (yes, I am skipping c) equal to your constant C, and p_a, p_b etc all prime.
No, that looks wrong. Take c = 2, for example. You seem to be saying that p^2 ought to be a solution. But D(p^2) = 2p (because D(p^2) = D(p) p + p D(p) = p + p = 2p).
Although you might also want to consider that "integration" (really, indefinite integrals, AKA antiderivative) is only defined up to a constant. So why couldn't it be the same for this "number derivative"? Perhaps the "antiderivative" is only defined up to something. It'd be a fun exercise, if you're interested. Can you figure out under what conditions do you get D(a) = D(b)? Put differently, given an integer c, what are the solutions to D(x) = c?