The solutions are the ones you listed.
The solutions all have shape $y=(x-c)^2$ or $y=0$. Thus if $b<0$, then none of the solutions curves pass through $(a,b)$. So for all pairs $(a,b)$ such that $b<0$, there cannot be a solution satisfying $y(a)=b$. We do not know (yet) whether these are all the pairs $(a,b)$ for which there is no solution, but soon we will.
For any $a$, if $b=0$ there are exactly two solutions satisfying $y(a)=b$, the singular solution and the solution $y=(x-a)^2$.
Finally, we look at pairs $(a,b)$ with $b$ positive. We look for values of $c$ such that $y(a)=b$.
The solution $y=(x-c)^2$ passes through $(a,b)$ if and only if $(a-c)^2=b$. This equation has exactly two solutions, $c=a\pm\sqrt{b}$.
Conclusion: (a) The pairs $(a,b)$ for which there is no solution satisfying $y(a)=b$ are all $(a,b)$ with $b<0$. (b) There are no pairs $(a,b)$ for which there are infinitely many solutions with initial condition $y(a)=b$. (c) For all remaining pairs $(a,b)$, that is, all pairs with $b \ge 0$, there are finitely many solutions, indeed exactly two solutions that satisfy $y(a)=b$.
The geometry: The conclusion can also be reached geometrically, by visualizing the family of parabolas. All of your parabolas are obtained by sliding the standard parabola $y=x^2$ along the $x$-axis. For any $(a,b)$ with $b \gt 0$, there are exactly two such parabolas that pass through (a,b): one whose "left" half goes through $(a,b)$, and one whose "right" half goes through $(a,b)$.
Note: One could interpret the word "finite" to include the possibility of $0$ solutions: $0$ is certainly finite! That is obviously not the intended interpretation here. But if we interpret "finite" as including $0$, the answer to (c) is all pairs $(a,b)$.
Best Answer
Use a solution with a general approach to compare your solution against.
Multiply the equation with $y$ to get $$ y^2 = (2py)x+\frac14(2py)^2. $$ This is a Clairaut equation for $v=y^2$, $$ v=v'x+\frac14(v')^2. $$ The singular solution in $v$ is given via the equation $0=x+\frac12v'$, thus $v'=-2x$ and directly inserted $$ v=-x^2. $$ As the part of that with non-negative $v$ is just a point, there is no way to observe the tangency of any curve to it when resolved to the original $y$.
So indeed all you calculated and observed is true.