Consider the case $a > |b|$, so the solutions are unbounded. The equation is separable, and we get implicit solutions of the form
$$ t = \int_{x_0}^x \frac{ds}{a - b \sin(s)} $$
which we can expand in a series in $1/a$ (uniformly convergent in $s$). Absolute convergence justifies interchanging sum and integral, so (for fixed $x$)
$$\eqalign{ t &= \int_{x_0}^x ds\; \sum_{k=0}^\infty (-b \sin(s))^k a^{-1-k}\cr &= \frac{x-x_0}{a} + \sum_{k=1}^\infty a^{-1-k}(-b)^k \int_{x_0}^x \sin^k(s)\; ds\cr &= \frac{x-x_0}{a} + O(a^{-2})}$$
Thus $x - x_0 = a t + O(a^{-1})$, which I believe is what you meant.