Yes. Scale $A$ by a positive factor and we may assume that $\max_ia_{ii}<1$. Then $B:=I-A$ is positive and $$ \sum_j|b_{ij}| =|b_{ii}|+\sum_{j\
e i}|b_{ij}| =1-a_{ii}+\sum_{j\
e i}|a_{ij}| <1 $$ for each $i$. Hence $\|B\|_\infty<1$ and we may expand $A^{-1}=(I-B)^{-1}$ as an infinite sum $I+B+B^2+\ldots$. However, as $B$ is positive, the infinite sum is positive too.