In the two dimensional case there is a simple proof. consider the set $U = \{ ax-by = 0 : x,y\in\mathbb{N}\}$. Let $A = (a,-b)$ and $v \in \mathbb{N}^2$, then for any $u\in U$, $A\cdot (v + u) = A\cdot v$. Furthermore, we have that, since $A\in\mathbb{Z}^2$, the inner product of $A$ against any vector in $\mathbb{N}^2$ is integral. So we can foliate $\mathbb{N}^2$ by lines $U_k$ such that $A\cdot u_k = k$ for $u_k \in U_k$.
Now,
$$ \sum_{U_k} \frac{1}{|A\cdot u| u_1 u_2} = \frac{1}{|k|} \sum_{U} \frac{1}{(x_k + u_1)(y_k + u_2)} $$
where to get to the right hand side we used the fact that on each $U_k$, as long as $a,b > 0$, we have a linear ordering of the elements by distance to the origin, and we can pick $x_k, y_k$ to be the infimum. Now, as long as $a,b \neq 0$, we have that $u_1,u_2$ are both increasing, so the inner sum is summable. Furthermore, a little bit of geometry shows you that
$$ \sum_U \frac{1}{(x_k + u_1)(y_k + u_2)} < \frac{2}{\sqrt{x_ky_k}} \sum_U \frac{1}{u_1u_2} $$
(Edit for clarification: Using the fact that $(u_1,u_2)$ are integer multiples of a unique vector $(\hat{u}_1,\hat{u}_2)$, you have
$$ \sum_U \frac{1}{(x_k + u_1)(y_k + u_2)} \leq \frac{1}{x_ky_k}\sum_j \frac{1}{1 + j^2\frac{\hat{u}_1\hat{u}_2}{x_ky_k}} $$
[here we also need to use that $\hat{u}_1\hat{u}_2 \neq 0$] then you use the fact that
$$ \sum_j \frac{1}{1 + \alpha j^2} \leq 1 + (\alpha)^{-1/2} \sum_k \frac{1}{1 + k^2}$$
which I leave as an exercise [if you run out of ideas, you can always prove this using integral comparison].)
The last thing to note is that $\frac{1}{x_ky_k} = O(|k|^{-1})$ using that the points $(1,k)$ and $(k,1)$ are the infimum points for the slices $U_{-bk}$ and $U_{ak}$ respectively. So
$$ \sum_{k\neq 0} \sum_{U_k} \frac{1}{|A\cdot u| u_1u_2} \leq \sum_{k\neq 0} \frac{C}{k^{3/2}} \sum_U \frac{1}{u_1 u_2} < \infty$$
With some addition condition on what $A$ you allow, I think the claim may also be true in higher dimensions. I'll add a proof to that effect if I can think of it.