I believe this problem can be tackled using polar coordinates. The equation of an conic in these coordinate is:
$$r(\theta) = \frac{r_0}{1+e \cos (\theta + \phi)},$$
where $r$ is the distance to one of the foci of the ellipse (the center of mass of the system, or the center of the star if it is much more massive than the orbiting object), $e$ is the excentricity and $\theta$ is chosen for instance so that $\theta = 0$ for your initial position (not necessarily one of the axis of the ellipse). For $e < 1$, the trajectory is closed and corresponds to an ellipse. When $e \geq 1$, the planet escapes to infinity (parabole or hyperbole), but I think that the following stays true (you just have to be careful about the allowed $\theta$'s).
Now what you need to do is convert your "target" from $x$, $y$ coordinates to $r_{\mathrm{target}}$, $\theta_{\mathrm{target}}$. Then you need to find the relation between your initial angle $\alpha$ (fixed), initial velocity $v_0$ with $r_0$, $e$ and $\phi$. I have not worked out the details but I assume this can be done using the definition of excentricity and inverse trigonometric functions. I will try to look into it. Now that you have $r_0(\alpha, v_0)$, $e(\alpha, v_0)$ and $\phi(\alpha, v_0)$, you can just inject into the ellipse equation and look at $\theta = \theta_{\mathrm{target}}$. You need that
$$r_{\mathrm{target}} = r(\alpha, v_0, \theta = \theta_{\mathrm{target}}) = \frac{r_0(\alpha, v_0)}{1 + e(\alpha, v_0) \cos(\theta_{\mathrm{target}} + \phi(\alpha, v_0))}.$$
I do not know if this is the easiest way to do it, but I believe it can be done nonetheless.
If you want to check if a solution exists, you need to wonder what happens if, given an angle, you increase progressively the velocity. For low velocities, you effectively fall rapidly towards the star. For very large velocity, you go through a more and more elongated ellipse. If you go beyond the escape velocity, you will not follow a closed trajectory anymore but escape to infinity (parabole or hyperbole). If your velocity becomes infinite, you basically go into a straight line. For all the velocities in between, you basically span all the half-plane separated by a straight line passing through your initial position and parallel to your initial velocity, and containing the star.
On the above picture, the green part of the plane are the accessible positions. You see for instance that if you would want to hit the target shown here, you would need a velocity $v_0$ between the one represented in blue (ellipse) and the one represented in red (hyperbole). The red part of the picture, on the other hand, correspond to the points that cannot be accessed using this value of $\alpha$.
If you do find a solution and want to check whether or not you would go through the star at some point, you just need to check if the minimal distance which is $r_0/(1+e)$ is larger or smaller than the radius of the star (assuming the star is very massive and the foci of the ellipse are merged with the center of the star).
Hope this helps!
Preliminary calculations :
the eccentricity is defined as :
$$e = \sqrt{1 + \frac{2 \varepsilon h^{2}}{\mu^2}},$$
where $\varepsilon$ is the specific orbital energy (total energy divided by the reduced mass), $\mu$ the standard gravitational parameter based on the total mass, and $h$ the specific relative angular momentum (angular momentum divided by the reduced mass) (-> https://en.wikipedia.org/wiki/Orbital_eccentricity).
Assuming a large star mass so that the reduced mass $\mu$ is just the mass of the planet $m$ and the total mass is roughly the mass of the star $M$, you find:
$$e = \sqrt{1 + \frac{2 (v_0^2 - \frac{2GM}{d}) d^2 v_0^2 \cos^2 (\alpha)}{G^2 M^2}},$$
where $d$ is the distance from the star to the planet initially, and $\alpha$ is defined so that $\alpha = 0$ if the velocity is purely radial initially. By the way, this gives you a second order equation in $v_0^2$, which allows you to invert it (although I am not sure if you want to invert the equation at this point...).
Then, to find $\phi$, you can use the Laplace-Runge-Lenz vector (https://en.wikipedia.org/wiki/Laplace%E2%80%93Runge%E2%80%93Lenz_vector), defined as $\overrightarrow{A} = \overrightarrow{p} \times \overrightarrow{L} - GMm^2 \overrightarrow{e_r}$, which indicates the direction for which $\theta + \phi = 0$. If you choose your x-axis to line up with your initial position, you find that $\overrightarrow{A} = m^2 v_0^2 d (\cos^2(\alpha) \overrightarrow{e_x} + \cos(\alpha) \sin(\alpha) \overrightarrow{e_y}) - GMm^2 \overrightarrow{e_x}$ (please check again).
From this you should be able to find $\phi$ and then $r_0$. Good luck!
