Step I. Schwarz Reflection Principle implies that $f(\overline{z})=\overline{f(z)}$. Hence, if $\,\mathrm{Im}\, z>0,$ then $\mathrm{Im}\, f(z)>0.$
Step II. $f'(x)\ne 0$, for all $x\in\mathbb R$.
Assume that $f'(x_0)=0$, for some $x_0\in\mathbb R$. Then $g(x)=f(x)-f(x_0)$ satisfies whichever properties $f$ is required to satisfy in the OP. Now, $g(x_0)=g'(x_0)=0$, and thus $\frac{1}{2\pi i}\int_{|z-x_0|=r}g'/g\ge 2$, where $r>0$ is chosen so that $g(x_0-r), g(x_0+r)\ne 0$. But the total argument change of $g'/g$ can not exceed $\pi$ in each of the two half planes.
Step III. $f$ vanishes at unique $x_0\in\mathbb R$. Uniqueness follows from Step II. If $f(x)\ne 0$, for all $x\in\mathbb R$, then $f(z)\ne 0$, for all $z\in\mathbb C$, and hence, the exists an entire $g$, such that $f=\mathrm{e}^g$. Since $\,\mathrm{Im}\, z>0,$ implies $\mathrm{Im}\, f(z)>0,$ then $\,\mathrm{Im}\, z>0,$ implies $\mathrm{Im}\, g(z)\in \big(2\pi k^+, 2\pi(k^++1)\big),$ for some integer $k^+$, and similarly
$\,\mathrm{Im}\, z<0,$ implies $\mathrm{Im}\, g(z)\in \big(2\pi (k^--1), 2\pi k^-\big)$, for some integer $k^-$. In which case $\mathrm{Im}\,g$ would be bounded, and so would be $g$. Without loss of generality, $x_0=0$.
Step III implies that $f$ is of the form $f(z)=(z-x_0)\mathrm{e}^{g(z)}$.
In the upper half plane, $f(z)=\mathrm{e}^{g(z)+\log (z-x_0)}$.
(Here $\log (z-x_0)$ is the branch, defined in $\mathbb C\setminus (-\infty,x_0],$ with $\mathrm{Im}\,\log(z-x_0)\in (0,\pi)$, when $\mathrm{Im}\,(z-x_0)>0$.) Again, $\mathrm{Im}\,f(z)>0$, implies that there exists an integer $k$ such that
$\mathrm{Im}\,\big(g(z)+\log (z-x_0)\big)\in \big(2k\pi,(2k+1)\pi\big)$, and since $\mathrm{Im}\,\log (z-x_0)\in (0,\pi),$ then $\mathrm{Im}\,g(z)\in \big((2k-1)\pi,(2k+1)\pi\big)$. Similarly, in the lower half plane, $\mathrm{Im}\,g(z)\in \big((2k'-1)\pi,(2k'+1)\pi\big)$, for some integer $k'$. Thus $g$ is bounded and hence constant, which implies that $f(z)=a(z-x_0)$, for some $a\ne 0$ real.
