As I already noticed in my comment, you can find a partial answer to your question and further details in my other answer and the references that were provided.
What you seem to be asking, is "how do we know that the average of probability forecasts is a valid probability?", at least this is how I understand it. Your question asks about taking averages of multiple probabilistic forecasts to take pooled forecast, so it is closely resorted to linear opinion pools (Stone, 1961).
First thing to notice is that a probability forecast, is in fact a conditional probability distribution. Taking arithmetic mean is a special case of taking a weighted sum $\sum_k w_k x_k$ with $\forall\, w_k > 0$ and $\sum_k w_k = 1$, where $w_1 = w_2 = \dots = w_n = n^{-1}$, so it is a convex combination. A weighted sum of probability distributions leads to a mixture distribution
$$
p(x) = \sum_k w_k \,p_k(x)
$$
where $p_k$ are some probability density (or mass) functions.
As already said by Cowboy Trader, you can think of this in terms of basic laws of probability. Given the properties of weights $w_i$, we can think of them as of probabilities, the most meaningful interpretation would be considering them as prior probabilities for choosing those forecasts. In such case, their joint distribution is
$$
p(x, k) = p_k(x) \,w_k = p(x|k) \, p(k)
$$
what follows from the definition of conditional probability. When we have joint distribution, we can calculate marginal distribution of it by the law of total probability
$$
p(x) = \sum_k p(x, k) = \sum_k p(x|k) \, p(k)
$$
If you also want to ask "why do people use it?", then the answer is: because it just works.