I want to find all entire functions $f$ such that for all $z \in \mathbb{C}$, $u(x, y) = av(x, y) + b$ for $a, b \in \mathbb{R}$. My attempt is as follows.$\DeclareMathOperator{\Re}{Re}\DeclareMathOperator{\Im}{Im}$
If $u(x, y) = av(x, y) + b$, then $u(x, y) - av(x, y) - b = 0$. Since $\Re(f(z)) = u(x, y)$ and $\Im(f(z)) = -\Re(if(z)) = v(x, y)$, then it follows that \begin{equation*} \Re(f(z)) - a\Re(if(z)) - b = 0 \end{equation*} Therefore, \begin{align*} \Re(f(z) - aif(z) - b) &= 0 \\ f(z) - aif(z) - b &= iC, \exists C \in \mathbb{R} \\ f(z)(1 - ai) &= iC + b \\ f(z) &= \dfrac{b + iC}{1 - ai} \end{align*} I am not sure if this is the right approach, and if not, how can I improve finding all entire functions satisfying the conditions? Appreciate some help or techniques.
Your $C$ depends on $z$ so you have not solved the problem.
By Cauchy Riemann equations we get $v_y=u_x=av_x$ and $-v_x=u_y=av_y$. Hence, $v_y=av_x=-a^{2}v_y$. So $(1+a^{2})v_y=0$ If $a^{2} \neq -1$ (i.e. $a \neq \pm i$) this gives $v_y=0$. But $av_x=v_y$ so $v_x=0$ if $a \neq 0$. In the case $a \neq 0, a \neq \pm i$ we have proved that $v_x=v_y=0$ so $v$ is a constant (and so is $u$). In this case $f$ is necessarily a constant function.
The cases, $a=0, a=i, a=-i$ are left to you.
EDIT: It is given that $a$ is real. So only the case $a=0$ is left to you. Thanks to Samuel M. A. Luque for pointing this out.