Consider the possible corners that a bishop can travel to, depending on the color pattern on the chessboard. Then, if one can get to a corner $(x',y')$, then

$x\equiv x'\pmod{a}$, and $y\equiv y'\pmod{b}$ $ ~~~~$(*)

The reason is that all four possible moves preserve the $x$ coordinate $\pmod{a}$, and the $y$ coordinate $\pmod{b}$. So in order to get to $(x',y')$, the conditions of (*) must both hold. (similarly for the other corner).

Now suppose that the conditions hold. Then $x-x'=sa$ and $y-y'=tb$, for some integers $s,t$. Note that the four moves all preserve the parity of $m+n$. The final position has $s+t=0$, so if the initial position has $s+t$ odd, the position can't be solved. If instead the initial position has $s+t$ even, then the position can be solved, using $\max(|s|,|t|)$ moves, as follows. Without loss suppose that $|t|\ge |s|$, and that $t>0$. Each move will decrease $t$ by one. Initially, we move $s$ to approach $0$; once it gets there then we alternate moving $s$ back one step, then again to $0$, until $t$ finally arrives at $0$.

Worked example, as requested, for $n=5, m=7, i=1, j=3, a=2, b=2$

Suppose we color the board so that the bishop begins on a black square. As it happens, all four corners are black. Considering the $(1,1)$ corner, we have $i-1=0a$ and $j-1=1b$. Since $0+1$ is odd, we cannot reach that corner. Considering the $(7,1)$ corner, we have $i-7=-3a$ and $j-1=1b$. Since $-3+1$ is even, we can reach the corner. It takes 3 steps, where we increase the x-coordinate at each step. The y-coordinate first goes down (to its minimum), then back up, then down again. Considering the $(1,5)$ corner, we have $i-1=0a, j-5=-1b$. Since $0+(-1)$ is odd, we cannot reach that corner. Lastly, considering the $(7,5)$ corner, we have $i-7=-3a, j-5=-1b$. Since $-3+(-1)$ is even, we can reach the corner, again in 3 steps.