The standard parametrization to get the Kepler rules is
$$
z=r+i(s-1)=\frac{e^{i\phi(t)}}{u(\phi(t))}.
$$
To incorporate the velocity data we need the first derivative of this parametrization,
$$
\dot z=e^{i\phi}\left(\frac{i}{u}-\frac{u'}{u^2}\right)\dot\phi=z\left(i-\frac{u'}{u}\right)\dot \phi.
$$
Per the naming conventions of the Kepler laws you then get the first law as
$$
u(\phi)=\frac{1+e\cos(\phi-\phi_{peri})}R\text{ with }R=\frac{L^2}{G}\tag{K1}
$$
and the second law as
$$
\dot\phi=Lu(\phi)^2\tag{K2}
$$
From your position data you have
- $0+i(0-1)=1e^{-i\pi/2}$, so $u(-\frac\pi2)=1$, and
- $1/2+i(1-1)=\frac12e^{i·0}$, so $u(0)=2$.
The velocity data point gives at $\phi=-\frac\pi2$, $u=1$
$$
1+i·0=(-i)\left(i-u'(\phi)\right)\dot \phi
$$
which allows to identify $\dot\phi=1$, $u'=0$, making this point the apoapsis. As the periapsis is opposite this, $\phi_{peri}=\frac\pi2$. The area velocity now follows as $L=1$.
In view of the first Kepler law, the positions now give the equations
\begin{align}
R&=1-e\\
2R&=1+e·0
\end{align}
giving $R=e=\frac12$ and thus $G=\frac{L^2}{R}=2$ exactly, as you had found numerically.
The period of the orbit is $T=2\pi ab$, where $a$ and $b$ are the long and short half-axis of the orbit $3r^2+\frac94(s-\frac23)^2=1$. Thus $T=2\pi·\frac23·\frac1{\sqrt3}=\frac{4\pi}{3\sqrt3}$.
Details of the derivation of the Kepler laws
Inserting the second derivative of $z$ into the gravity equation gives
\begin{align}
-\frac{Gz}{|z|^3}=\ddot z&=z\left(i-\frac{u'}{u}\right)^2\dot \phi^2
+z\left(-\frac{u''}{u}+\frac{u'^2}{u^2}\right)\dot \phi^2
+z\left(i-\frac{u'}{u}\right)\ddot \phi,
\\
\implies
-Gu^3&=\left(i-\frac{u'}{u}\right)\left(\ddot\phi-2\frac{u'}{u}\dot\phi^2\right)+\left(-1-\frac{u''}{u}\right)\dot\phi^2
\end{align}
As the imaginary part on the left is zero, so it has to be on the right, giving the second Kepler law
$$
\frac{\ddot\phi}{\dot\phi}=2\frac{u'(\phi)\dot\phi}{u(\phi)}\implies\dot\phi=Lu(\phi)^2.
$$
What remains is the first Kepler law,
$$
\frac{G}{L^2}=u''+u
$$
with its solution parametrized as in (K1)
Best Answer
Let $g(t,x) = \exp(-t x) \left(1-t^2\right)^{-1/2}$. Note that $$ \frac{\partial^2}{\partial x^2} g(t,x) + \frac{1}{x} \frac{\partial}{\partial x} g(t,x) - g(t,x) \stackrel{\text{evaluate}}{=} \exp(-t x) \frac{-t-x(1-t^2)}{\sqrt{1-t^2}} = \frac{\partial}{\partial t} \left( \exp(-t x) \sqrt{1-t^2} \right) $$ Now $f(x) = \int_1^\infty g(t,x) \mathrm{d}t$ giving: $$ f^{\prime\prime}(x) + \frac{1}{x} f^\prime(x) - f(x) = \left. \exp(-t x) \sqrt{1-t^2} \right|_{t \to 1^+}^{t \to \infty} $$ The right-hand-side vanishes for $\Re(x)>0$, which is incidentally the same condition when $\int_1^\infty g(t,x)\,\mathrm{d}t$ converges, thus establishing the claim.
By the way, function $K(x)$ defined by the integral is the modified Bessel function of the second kind of order $0$.