How can be proved that the softmax output forms a probability distribution and the sigmoid output does not?
Softmax maps $ f:ℝ^n\rightarrow (0,1)^n$ such that $\sum f(\vec x) =1$. Therefore, we can interpret the output of softmax as probabilities.
With sigmoidal activation, there are no such constraints for summation, so even though $ 0&lt;S(\vec x)&lt;1$, it is not guaranteed that $\sum S(\vec x)=1$. The sigmoidal function does not normalize the outputs, so in your example where class 0 has output $0.7$, class 1 could have any value in $(0,1)$, which might not be $0.3$.

Here's an example:
$\vec x=[-5,\pi,\frac{1}{3},0] $
$ f(\vec x)\approxeq [2.6379\times10^{-4},0.9059,0.05464]$
$ S(\vec x)\approxeq [6.693\times10^{-3},0.9586,0.5826,0.5] $
Because $0&lt;f(\vec x)&lt;1$ and $\sum f(\vec x)=1$, the softmax output vector can be interpreted as probabilities. On the other hand, $ \sum S(\vec x) &gt; 1$, so you cannot interpret the sigmoidal output as a probability distribution, even though $ 0&lt;S(\vec x)&lt;1$
(I chose the above $\vec x$ arbitrarily to demonstrate that the inputs need not be negative, non-negative, rational, etc., hence $\vec x\in ℝ^n$)