If you don't consider atmospheric drag and other dissipative effects, assuming that the Earth is non-rotating, and ignore general relativity (which is a lot of assumptions to make, and later I can try to edit this to comment on how they might be relaxed), a ballistic trajectory is just a segment of an orbit. So finding the launch parameters just entails finding an orbit that intersects both the launch point and the destination.
Orbital trajectories are given by the equation
$$r_o = \frac{\rho}{1 + \epsilon\cos\phi}$$
where $\rho$ is a length parameter related to the size of the orbit and $\epsilon$ is the eccentricity. At the points where the orbit intersects the Earth's surface, you'll have $r_o$ equaling the radius of the Earth at that point; let's say $r_o = R_L$ for the launch point and $r_o = R_D$ for the destination. This gives you the following two equations,
$$R_L = \frac{\rho}{1 + \epsilon\cos\phi_L}$$
$$R_D = \frac{\rho}{1 + \epsilon\cos\phi_D}$$
along with the constraint that $\phi_D - \phi_L$ (or vice-versa, depending on how you define coordinates) has to be equal to the angular separation $\Delta$ between the source and destination. That angular separation can be calculated using the law of haversines.
This system of equations can be solved for the orbital parameters as follows, assuming $R_D \neq R_L$ (see below for the other case),
$$\epsilon = \frac{R_L - R_D}{R_D\cos(\phi_L + \Delta) - R_L\cos\phi_L}$$
$$\rho = \frac{R_D R_L [\cos(\phi_L + \Delta) - \cos\phi_L]}{R_D\cos(\phi_L + \Delta) - R_L\cos\phi_L}$$
Notice that they are dependent on $\phi_L$; this is because only the difference $\phi_D - \phi_L = \Delta$ is physically relevant, so you're free to choose the actual values to be whatever you like. Your choice of $\phi_L$ will influence the shape of the trajectory.
Having calculated $\epsilon$ and $\rho$, you can determine the launch angle $\alpha$ above the horizon by calculating the slope of the orbit at the launch point:
$$\tan\alpha = \left.\frac{1}{r_o}\frac{\mathrm{d}r_o}{\mathrm{d}\phi}\right|_{\phi_L} = \frac{\epsilon\sin\phi_L}{1 + \epsilon\cos\phi_L} = \frac{(R_L - R_D)\sin\phi_L}{R_D[\cos(\phi_L + \Delta) - \cos\phi_L]}$$
To determine the speed, you can use the fact that (according to my notes) the total energy of the projectile is given by
$$E = \frac{GMm(\epsilon^2 - 1)}{2\rho}$$
which is equal to the sum of kinetic and potential energies, $\frac{1}{2}mv^2 - \frac{GMm}{r}$. ($m$ is the mass of the projectile, $M$ is that of the Earth) Plugging in $r = R_L$, I get
$$v = \sqrt{\frac{2GM}{R_L} + \frac{GM(\epsilon^2 - 1)}{\rho}}$$
So the bottom line is that you plug $R_D$, $R_L$, $\Delta$, and some choice of $\phi_L$ into the formulas for $v$ and $\alpha$ to get the launch parameters.
I mentioned that the procedure above hits a snag if $R_D = R_L$. You'd wind up calculating $\epsilon = 0$, which on a spherical planet entails rolling your projectile along the ground ;-) which doesn't make sense.
If $R_D = R_L$, you can go back to the orbital equations and find that $\cos\phi_D = \cos\phi_L$. (Alternatively, this could come from the condition that the denominator of $\epsilon$ also be zero.) The only way to satisfy this is by setting $\phi_L = -\phi_D = \Delta/2$. This gives you the condition
$$\rho = R_L\left(1 + \epsilon\cos\frac{\Delta}{2}\right)$$
Again, you have a degree of freedom: you can choose any value of $\epsilon$, and plugging into this equation will give you the corresponding value of $\rho$. Once you have those, you can use the same procedures to calculate the launch angle and speed:
$$\tan\alpha = \left.\frac{1}{r_o}\frac{\mathrm{d}r_o}{\mathrm{d}\phi}\right|_{\phi_L} = \frac{\epsilon\sin\frac{\Delta}{2}}{1 + \epsilon\cos\frac{\Delta}{2}}$$
$$v = \sqrt{\frac{2GM}{R_L} + \frac{GM(\epsilon^2 - 1)}{R_L(1 + \epsilon\cos\frac{\Delta}{2})}}$$