This scenario is explicitly handled by Gordan's theorem, which states
$$
\text{either} \quad
\exists x \in \mathbb{R}_+^m\setminus\{0\} \centerdot Ax = 0,
\quad\text{or}\quad
\exists y\in\mathbb{R}^n\centerdot A^\top y > 0,
$$

where $\mathbb{R}_+$ denotes nonnegative reals.
(Like Farkas's Lemma, this is a "Theorem of Alternatives"; furthermore, it can be proved from Farkas's lemma.)

A nice way to prove this is, as in Theorem 2.2.6 of Borwein & Lewis's text "Convex Analysis and Nonlinear Optimization", to consider the related optimization problem
$$
\inf_y \quad\underbrace{\ln\left(\sum_{i=1}^m \exp(y^\top A \textbf e_i)\right)}_{f(y)};
$$
as stated in that theorem, $f(y)$ is unbounded below iff there exists $y$ so that $A^\top y > 0$. As such, this also gives an unconstrained optimization problem you can plug into your favorite solver to determine which of the two scenarios you are in. Alternatively, you can explicitly solve for either the primal variables $x$ or the dual variables $y$ by considering a similar max entropy problem (i.e.
$\inf_y\sum_i \exp(y^\top A\textbf{e}_i)$, which approaches 0 iff the desired $y$ exists) or its dual (you can find this in the above book, as well as papers by the same authors).

Anyway, considering Gordan's theorem, your condition on the columns (which can be written $\textbf{1}^\top A = 0$) has no relationship to the question at hand. In one of your comments you mentioned wanting to generate these matrices. To pick positive examples, fix a satisfying $x$, and construct rows $b_i'$ by first getting some $b_i$ and setting $b_i' := b_i - (x^\top b_i)x / (x^\top x)$; to pick negative examples, by Gordan's theorem, choose some nonzero $y$, and then consider adding to $A$ a column $a_i$, including it if it satisfies $a_i^\top y > 0$.