Arithmetician (Ph.D. in number theory) here. This answer is completely correct.
As another reason why the factorization of 1 should be the empty set: suppose you have two positive integers m and n. Write S(m) and S(n) for their sets of prime divisors, counted with multiplicity.
Then S(mn) is the union of S(m) and S(n). We need S(1) to be the empty set to make this rule consistent.
Analogous to how log(1) is equal to 0.
The analogy is a good one, as zero is the additive identity (https://en.wikipedia.org/wiki/Additive_identity) of the integers, and one is the multiplicative one (https://en.m.wikipedia.org/wiki/1#Mathematics)
I'm not saying it's incorrect, but you have to explain that edge case away. At the same time allowing 1 to be prime would remove that problem given that e.g. 6 = 2^1 * 3 ^1 and 6 = 1^81 * 2^1 * 3^1 are the same unique factorisation because, well, monoids.
Edit: `s/fields/monoids`
Now you are the one squirming.
Cleanly accounting for edge cases isn't squirming, it's logical thinking.
The fundamental theorem doesn't have to state that the prime factorization is nonempty.
It's true that some mathematicians have defined 1 as a prime number and there's nothing logically inconsistent about doing so, but it makes most theorems and formulas in number theory more complex and so this definition has fallen out of favor.
Edit: I think the Wikipedia article on the empty product gives some quite nice examples of the benefits of a closely related concept. https://en.wikipedia.org/wiki/Empty_product