For a non-constant transcendental entire function $f$ and distinct constants $a$ and $b,$
suppose that $f(z)-az$ and $f(z)-bz$ are periodic with periods $x$ and $y$ respectively.
I have to show that $f(z)-az$ and $f(z)-bz$ are both bounded on a straight line $\{tx: t\in (-∞ , +∞)\}$.
Attempt: since $f(z)-az$ and $f(z)-bz $ are periodic with periods $x$ and $y$ respectively, we find that $f'(z)$ has periods $x$ and $y$ and hence either $x/y$ is a real number or $f'(z)$ is a constant. But I can reject the latter case since $f(z)$ is given to be transcendental. But my problem is how this leads to the conclusion that $f(z)- az$ and $f(z)- bz$ are both bounded on a straight line $\{tx: t\in (-∞ , +∞)\}$.
I am stuck here.
As you observed, the assumptions imply that $f'$ is periodic with periods $x$ and $y$. For a non-constant entire function, the only possibilities for the set of periods is either $\{0\}$ or $\omega \mathbb{Z}$, where $\omega \ne 0$ is some primitive period. In particular, this implies that $x=k\omega$ and $y=l \omega$ with some non-zero integers $k$ and $l$. Then $p = kl \omega$ is a common period of $f(z)-az$ and $f(z)-bz$, which implies that it is also a period of their difference $(b-a)z$. However, since this is not a periodic function, there is no such $f$.