In the example linked, the transfer orbit is elliptical about the Sun, but to get there you must hyperbolically escape the Earth. This is a patched conic approximation, where the multiple conic sections/orbits (about different gravitationally significant bodies) are 'patched' together at their intersections.
Also note that for the example linked the equation does not "calculate the needed velocity at the edge of Earth's SOI", this was done prior. The "needed velocity at the edge of Earth's SOI" is an input to the equation. The output is the velocity at a low Earth orbit height.
The equation is valid for all conic sections; however, with a negative sign convention, much like how the semi-major axis of a hyperbolic orbit is negative (but opposite). From Wikipedia (emphasis added):
The semi major axis ($a$) is not immediately visible with an
hyperbolic trajectory but can be constructed as it is the distance
from periapsis to the point where the two asymptotes cross. Usually,
by convention, it is negative, to keep various equations consistent
with elliptical orbits.
The semi major axis is directly linked to the specific orbital energy
($\epsilon$) or characteristic energy $C_{3}$ of the orbit, and to the
velocity the body attains at as the distance tends to infinity, the
hyperbolic excess velocity ($v_{\infty }$).
$v_{\infty }^{2}=2\epsilon =C_{3}=-\mu /a$, $v_{\infty }^{2}=2\epsilon
> =C_{3}=-\mu /a$, or $a=-{\mu /{v_{\infty }^{2}}}$, $a=-{\mu /{v_{\infty }^{2}}}$
Also note that there are some shenanigans with the negative sign convention to make the equation(s) hold, mainly that squaring/square rooting a value does not affect its sign (i.e., $\sqrt{-x} \to -\sqrt{x}$).
For a circular 250 km low Earth orbit, the $v_{\infty}^2$ would be:
$$v^2=v_e^2+v_{\infty}^2 \to v_{\infty}^2=v^2-v_e^2=\frac{\mu}{r}-\frac{2\mu}{r} \to v_{\infty}^2=-60km^2/s^2$$
Negative, as expected for an elliptical (circle is a special case of an ellipse) orbit.
