I'm still working on trying to prove the results here with while satisfying as many axioms of conditionals as possible, and ideally with countably additive probabilities. It's not all that easy. Anyway, while thinking about it, I came up with a method of proof that while probably not new, I have never used before.
Suppose we want to close some set A0 under some finite set of operations with finite arity. One way is to inductively form a sequence of larger and larger sets, A0⊆A1⊆A2⊆..., such that An+1 is something like the result of applying the operations to all the tuples of elements of An. Then take the union A of the An, and that will be our closure, because any finite collection of elements of A will all be in some An for a large enough n, and hence the result of applying the operations to that collection will be in An+1, and hence in A. So A will be closed under the operations.
But what if our operations have countably infinite arity? An example is probability spaces which we want to be closed under countable unions. In that case, the above method of proof may not be good enough. But there is a cool modification given the Axiom of Choice. By Choice, let β be an ordinal with uncountable cofinality. Use transfinite recursion over α < β to generate a sequence of larger and larger sets Aα, such that Aα is the result of applying the operations to the countable sequences of elements of Aα, and letting Aα, for α a limit ordinal, be the result of applying the operations to the countable sequences of elements of the union of the sets for α'<α. Let A be the union of the Aα. Then a countable sequence of elements of A will all be contained in some one of the Aα because of uncountable cofinality. And then the result of applying the operations to that sequence will be in Aα+1, and hence in A. So A will be closed under the operations.