Assume that $A \geq 0$, $(I-A)$ is invertible. I have shown that if we have a sequence $$ S_n = I + A + A^2 + \cdots $$ that $$(I-A)S_n = I - S^{n+1} = I$$ as $n \rightarrow \infty$.
How can I now show that $S_{n \rightarrow \infty} \geq 0$? "Positive" here means that all entries of the matrix are $\geq 0$.
Let $S = \lim_{n \to \infty} S_n$, where $S_n = I + A + \cdots + A^{n-1}$. Because $S_n$ converges, each entry of $S_n$ converges to the corresponding entry of $S$. That is, for each $i,j,$ $\lim_{n \to \infty} S_n(i,j) = S(i,j)$.
Now, if a sequence $x_n$ satisfies $x_n \to x$ with $x_n \geq 0$ for all $n$, then it must hold that $x \geq 0$. Conclude that each entry of $S$ is non-negative, which is to say that $S \geq 0$.