Context: I have come across this problem in the theory of anyons described by unitary braided tensor categories where the Frobenius-Perron dimension (quantum dimension) of a charge is significant for the charge being abelian or non-abelian.
The problem is the following: Given a matrix $N_a$ with only non-negative integer entries, the Perron- Frobenius theorem tells me that the eigenvalues of this matrix are bounded in absolute value by the Frobenius-Perron eigenvalue which I will call $d_a$. Now, given $d_a=1$, I want to proof that $$\sum_j (N_a)_{ij}=1$$ for all $i$. Together with the nonnegativity and the entries being integers, I deduce that this is equivalent to proving that the matrix has exactly one entry equal to 1 in every row and all the others equal to zero. In the Wiki article on the Perron-Frobenius theorem, I have seen the inequalities $$\min_i \sum_j(N_a)_{ij} \leq d_a \leq \max_i \sum_{ij} (N_a)_{ij} $$ but they do not seem to help me unfortunately.
It is also know that the corresponding eigenvector to $d_a$ has only positive entries which given that $d_a=1$ implies the matrix non-singular. I know from other sources that actually all elements of the eigenvector are even greater equal 1 but I would rather try to avoid using that if not needed. It might be needed however, if you think it is, please point out why. Furthermore, $N_a$ cannot have zero-columns, at least one element in every column must be non-zero
For convenience, allow me to write your $N_a$ as $N$.
Suppose $v$ is an eigenvector for the Perron eigenvalue $1$ with strictly positive entries. We have $\sum_{j=1}^n N_{ij} v_j = v_i$ for each $i$.
Since the entries of $N$ are nonnegative integers, we must have $N_{ij} = 0$ if $v_j > v_i$. Thus the only possible nonzero entries in column $j$ are in rows $i$ where $v_i \ge v_j$. Moreover if $v_j = v_i$ and $N_{ij} > 0$, the only possibility is $N_{ij} = 1$ and all other entries in row $i$ are $0$.
Now you say there are no $0$ columns. Thus for each $j$ we have some $i$ with $v_i \ge v_j$ and $N_{ij} > 0$. Define a mapping $f$ of $\{1,\ldots, n\}$ to itself so that $f(j)$ is some $i$ with $N_{ij} > 0$ ; in case there is a choice, take one with the greatest $v_i$. We have $v_{f(j)} \ge v_j$.
Now I claim $N_{ij} > 0$ only for $v_i = v_j$, so that $v_{f(j)} = v_j$. Suppose this is not so. Take $j$ with the greatest possible $v_j$ for which $v_{f(j)} > v_i$. For $v_k > v_j$, $v_{f(k)} = v_k$, so $N_{f(k),k} = 1$ and all other entries in row $f(k)$ are $0$. This says that $f(k) \ne f(k')$ for all $k' \ne k$: the map $f$ is injective from $S = \{k: v_k > v_j\}$ to itself. Since $S$ is finite, $f$ must be surjective on $S$ as well. In particular, $f(j) = f(k)$ for some $k$, but then it is impossible to have $N_{f(j),j} > 0$.
The conclusion is that $f$ is a permutation of $\{1,\ldots, n\}$, and $N$ is a permutation matrix. This does indeed have all its row-sums $1$.