In $[1]$ the following (paraphrased) claim is made:
Let $q\in L^1_{loc}([0,\infty);\mathbb{R})$, and suppose $\varphi$ and $\theta$ solve the one-dimensional Schrödinger equation \begin{equation} -u''(x;\lambda)+q(x)u(x;\lambda)=\lambda u(x;\lambda)~~~(x\in[0,\infty)), \end{equation} subject to \begin{equation} \left\{\begin{array}{cc} \varphi(0;\lambda)=0, &\varphi'(0;\lambda)=1, \\ \theta(0;\lambda)=1, &\theta'(0;\lambda)=0. \end{array}\right. \end{equation} Then, for each fixed $x$, we have that $\varphi(x;\cdot)$ and $\theta(x;\cdot)$ are entire functions of order $1/2$.
I understand from a previous question I asked that for entireness one needs the solution to obey a $\lambda$-independent initial condition (since otherwise specifying an i.c. non-analytic in $\lambda$ gives a counter-example), and $\varphi,\theta$ do exactly this.
Further than this I am fairly stumped, and in particular have no idea how to approach the claim on the order. Recall that the order of an entire function $f$ is \begin{equation} \rho:=\inf\{m\geq0~|~f(z)=O(\exp(|z|^m))\text{ as }|z|\rightarrow\infty\}. \end{equation}
Is it perhaps related to the Gel'fand-Levitan integral transformation, which allows one to write \begin{align} \varphi(x;\lambda)=\frac{\sin(\sqrt\lambda x)}{\sqrt\lambda}+\int_0^x K(x,t)\frac{\sin(\sqrt\lambda t)}{\sqrt\lambda}\text{d}t, \\ \theta(x,\lambda)=\cos(\sqrt\lambda x)+\int_0^x K(x,t)\cos(\sqrt\lambda t)\text{d}t? \end{align} Here $K$ satisfies \begin{align} K_{xx}&=K_{tt}+q(x)K(x,t), \\ 2\frac{\text d}{\text dx}K(x,x)&=q(x),~K_t(x,0)=0. \end{align}
$1$. Bennewitz, C., A proof of the local Borg-Marchenko theorem, Commun. Math. Phys. $218$, pp. $131$--$132$ ($2001$).
This is a little too long for a comment. I wonder if the standard Picard iteration would work. If you want a solution $u$ with $u(0)=A$ and $u'(0)=B$, then $$ \begin{align} u'(x) & = B+\int_{0}^{x}(q(x_1)-\lambda)u(x_1)dx_1, \\ u(x) & = A+Bx+\int_{0}^{x}\int_{0}^{x_2}(q(x_1)-\lambda)u(x_1)dx_1 dx_2 \\ & = A+Bx + \int_{0}^{x}(x-x_1)(q(x_1)-\lambda)u(x_1)dx_1. \end{align} $$