For fixed $a$ and $b$, after $m$ steps, the frog will be at $a + mb$ if you haven't guessed right yet. The set of pairs $(a, b) \in \mathbb{N}\times \mathbb{N}$ is countable, so you can enumerate it, say as $(a_0, b_0), (a_1, b_1), \ldots$. Now guess $a_m + mb_m$ at step $m$ and you are bound to guess right eventually.