This conjecture is correct. Take $K=e$, and let $\gamma\leq 1/4$; we will fix $\gamma$ later.
First we give a crude estimate of $c_0$.
Let $g(z)=\sum_{1}^\infty c_nz^n.$ Since $|c_n|\leq e^n$,
we obtain $|g(z)|\leq e^n$ for $|z|\leq 1/2$ by the trivial estimate. Then by Cauchy, $|f'(z)|=|g'(z)|\leq 4\cdot e^n,\; |z|\leq 1/4$.
Then $$|c_0-e^{-n}|\leq (1/4)\max_{[0,1/4]}|f'|\leq (1/4)4\cdot e^n=e^n,$$
so $|c_0|\leq 1.5\cdot e^n$. Therefore
$$|f(z)|\leq |c_0|+|g(z)|\leq 3\cdot e^n,\quad |z|\leq 1/2.$$
Now consider the subharmonic function
$$u(z)=\log|f(z)|$$
in the region $D=\{z:|z|<1/2,z\not\in E\},$
where $E=\{ z\in(0,1/2): |f(z)|\leq e^{-n}\}.$
By the Two constants Theorem,
$$u(0)\leq -n\omega(0,E,D)+n\omega(0,C,D)+\log 3,\quad\quad\quad (1)$$
where $C=\{ z:|z|=1/2\}$ and $\omega$ is the harmonic measure. Now, according to a theorem of Beurling
(Nevanlinna, Analytic functions, Chap. IV, section 84),
$\omega(0,E,D)\geq \omega(0,E_0,D_0),$ where
$E_0$ is the segment $[\gamma,1/2]$ and $D_0$ is the complement of this segment to the disk $|z|<1/2$.
It is easy to obtain an explicit formula for this $\omega(0,E_0,D_0)$, (see for example Nevanlinna's book),
but we only need the fact that it tends to $1$ when $\gamma\to 0$, which is evident. Since
$\omega(0,C,D_0)=1-\omega(0,E_0,D_0)$, we can fix $\gamma$ such that
$\omega(0,E_0,D_0)>\omega(0,C,D_0)$. Then
from the inequality (1) we conclude that
$$\log|c_0|=u(0)\to-\infty,\; n\to\infty.$$
This proves the result.
From the explicit expression of $\omega(0,E_0,D_0)$ we can obtain an explicit value of $\gamma$.
Remark. Beurling theorem is not necessary for mere existence of $\gamma$; it is only needed to obtain an explicit value. It is clear without Beurling that
$\omega(0,E,D)\to 1$ when $\gamma\to 0$, by an elementary "compactness argument".
Let
$$f(z)=\frac{e^{iz}-1}{iz}.$$
This function is in the Hardy class for any upper half-plane,
and has these properties: $f(0)=1,$ $f(2\pi n)=0$,
$$|f(z)|\leq C\frac{e^y+1}{|z|+1},$$
(this evidently holds for large and small $|z|$, therefore there is a constant $C$ so that this holds everywhere).
Since the $L^2$ norm has the property $\| f(kx)\|^2=\| f(x)\|^2/k,$ and the norm is invariant under real translations, the following function $g$ belongs to the Hardy spaces for upper half-planes:
$$g(z)=\sum_{n=0}^\infty f(n^2(z-a_n)),$$
where $a_n\in\{2\pi k: k\in {\mathbf{Z}}\}$.
Moreover, it is entire, if the sequence $a_n$ grows fast enough. (It follows from the estimate for $f$ that the series is uniformly convergent on any compact subset of the plane).
Now $g(a_n)=1$, so $h(z)=(g(z-i)-1)/(z-i)$, where has all properties that you required.
Some extra labor is needed to show that $h$ can be made of finite order.
Best Answer
There is a zero-free entire function bounded in every left half-plane, and such that $f-1$ is in $H^2$ in every left half-plane.
Let $\gamma$ be the boundary of the region $$D=\left\{ x+iy: |y|<2\pi/3, x>0\right\} .$$ Consider the function $$g(z)=\int_\gamma \frac{\exp e^\zeta}{\zeta-z}d\zeta,\quad z\in {\mathbf{C}}\backslash D.$$ The integral evidently converges and $g(z)=O(1/z)$ in ${\mathbf{C}}\backslash D.$ Now, $g$ has an analytic continuation to an entire function: deforming the contour to $\partial D_t$, where $D_t=\{ x+iy:|y|<2\pi/3, x>t\},\; t>0$ does not change $g$ in $D$, and shows that $g$ has an analytic continuation to ${\mathbf{C}}\backslash D_t$, and this is for every $t>0$, so $g$ is entire. Now $f(z)=e^{g(z)}$ is the desired function. If you want upper half-planes take $f(iz)$.
Remark. You can improve the estimate $g(z)=O(1/z)$. Evidently, $g$ has infinitely many zeros $z_1,z_2,\ldots$. Then $g_k(z)=g(z)/((z-z_1)\ldots(z-z_k))$ satisfies $g_k(z)=O(z^{-k-1})$ as $z\to\infty$ outside $D_t$.
Remark 2. This construction is standard in the theory of entire functions, see, for example, Entire function bounded at every line Sometimes this $g$ is called the Mittag-Leffler function.