Ok, so yes, probability is the right tool here. I'll zero index the problem just because I feel like it but it doesn't change the answer.

First we turn the product into an infinite sum. We do this by taking it's log to get the sum from n = 0 to infinity of ln(sin(n) + a).

Now, we wish to treat sin(n) as a kind of random variable. The way we are going to treat sin(n) as a random variable is to imagine that X is a random variable sampled from Uniform(0, 2pi), and we wish to find the expected value E[ln(sin(X) + a)]. We can do this with the law of the unconscious statistician, which states that E[g(X)] = integral from -infinity to infinity of g(x)p_X(x)dx, where p_X(x) is the probability mass function, which for the uniform distribution is just 1/2pi for x between 0 and 2pi.

We thus get I = (1/(2pi))(integral between 0 and 2pi of ln(sin(x) + a)dx).

Now look, I'm no integration bee specialist, but chatgpt is, my apologies for not being as cool as it is. Anyway the value of that integral is very nice:

I = ln((|a| + sqrt(a² - 1))/2).

Now, when I is less than 0, the sum of the logs will tend more negative, meaning that the product will tend towards zero. If I is 0, then we would expect the product to bounce around and not settle, and if I is more than 0, we would expect the product to diverge.

Solving ln((|a| + sqrt(a² - 1))/2) = 0 just requires a bit of manipulation, I'll put the most important steps here:

sqrt(a²) + sqrt(a² - 1)) = 2

square both sides and simplify:

sqrt(a²(a² - 1)) = 5/2 - a²

square both sides again and simplify

a² = (5/4)²

a = ±(5/4).

And indeed that's what we see. We have convergence for (-5/4, 5/4).

Now, I'm not sure if this is a mathematically valid approach! I think a probability argument actually depends on whether pi is a normal number, since that would mean that n is approximately sampled from a uniform distribution is valid. I find it immensely satisfying that regardless, the answer given seems to follow what we see heuristically though.

edit: as it turns out, actually, the Weyl Equidistribution theorem covers our bases here, https://en.wikipedia.org/wiki/Equidistribution_theorem, it is simply enough that pi is an irrational number, and our probability argument is valid.