Here's a somewhat intuitive explanation, though not the most elegant formally:
We group all formable numbers based on how many copies of $a$ we need (i.e., the value of $x$). Note that if $x > b$, we can replace $b$ copies of $a$ by $a$ copies of $b$, so we can assume $1 \leq x \leq b$. Then the formable numbers are:
$$0, b, 2b, 3b, \dots$$
$$a, a+b, a+2b, a+3b, \dots$$
$$2a, 2a+b, 2a+2b, 2a+3b, \dots$$
$$(b-1)a, (b-1)a+b, (b-1)a+2b, (b-1)a+3b, \dots$$
Since $a$ and $b$ are coprime, no two groups overlap. Now consider some interval $[n-b+1,n]$ where $n \geq (b-1)a$. Since the interval is exactly $b$ integers long, and each group contains members spaced exactly $b$ apart, it must contain a member of each group. Since there are $b$ groups, and no two groups overlap, every integer in the interval must belong to some group, i.e., every integer in the interval is formable.
Thus every integer greater than or equal to $(b-1)a-b+1 = ab-(a+b)+1$ is formable.