First of all, if $a$ or $b$ equals $1$, then the claim is false, so suppose they are both greater than $1$.
Suppose $(x_0,y_0)$ is one solution. Then we also have that $(x_0+kb,y_0+ka)$ is a solution for any integer $k$. (Do you see why?)
Choose the value of $k$ that makes $x=x_0+kb$ as small as possible while still positive; then $x<b$, or we could have chosen a smaller value of $k$. Then we have:
Divide by $b$ and see what you get. This establishes why we have $y<a$.
To see why $y>0$, all we need is that $ax>1$, which follows from $a>1$ and $x>0$.