Yes, I indeed did something different in the second comment. I also made a mistake (namely, I said or implied that the sum of probabilities over a set $$${x = a, b < y\leq m}$$$ is the probability to eventually end up on that edge, which is incorrect), which is probably why you noticed that this doesn't solve the original problem immediately. So below I write a correction.

So I assume you are interested in probabilities $$$p(x, y)$$$ to reach $$$(x, y)$$$, then we just need to add them over all $$$(x, y)$$$ s.t. $$$xy > k$$$ or something. We can represent this probability as $$$p(x, y) = p_h(x, y) + p_v(x, y)$$$, where $$$p_h(x, y)$$$ is the probability to eventually come to $$$(x, y)$$$ from the right, and $$$p_v(x, y)$$$ is the same but from the above. Let's calculate the sum of $$$p_h(x, y)$$$ over all good cells grouping them by columns and the sum of $$$p_v(x, y)$$$ grouping them by rows. Then it is true that the sum of $$$p_h(x, y)$$$ over $$${(x, y)\,\colon\,x = a, b < y\leq m}$$$ is what I wrote in the second comment (i.e. $$$1/(a+b)$$$). The solution becomes even easier: link.

You may be confused because the equation for the rows was something like $$$2 / (x + y) - 2 / (x + m)$$$ instead of a simple ratio like $$$1 / (x + y)$$$ as in our case. The reason behind the $$$1$$$ in the numerator is that we decomposed that probability into the horizontal one and the vertical one (and they turned out to be the same, so $$$2/c(c+1)$$$ became two $$$1/c(c+1)$$$'s). The reason why we don't have the $$$-2/(x+m)$$$ leftover is that we no longer distinguish the border cells from the inner ones. In the initial solution we had something like $$$1/i(i+1) + \ldots + 1/j(j+1) = 1/i - 1/(j+1)$$$, but here we add one $$$1/(j+1)$$$ from the boundary, which cancels out, thus leaving only $$$1/i$$$.

Regarding explaining the initial solution combinatorically, I am not ready to do this in any way except the one I just described, as I now think it is the most natural way to think about it.