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 $(a,b)$.
The goal here is to reduce the equation to something only in terms of $y,p,dp/dy$. If you can get $x$ to one side, it will go away after differentiation.
If you solve for $x$
$$ x = \frac{y - y^2p^3}{2p} = \frac{y}{2p}-\frac{y^2p^2}{2} $$
Then differentiate
$$ \frac{1}{p} = \frac{1}{2p} - \frac{y}{2p^2}\frac{dp}{dy} - yp^2 - y^2p\frac{dp}{dy} $$
Multiplying through by $2p^2$
$$ p + (y + 2y^2p^3)\frac{dp}{dy} + 2yp^4 = 0 $$
Then factor
$$ (1 + 2yp^3)\left(p + y\frac{dp}{dy}\right) = 0 $$
The equation is now solvable for $p = f(y)$
Best Answer
Consider the diagram
where the point of intersection is a point near the evelope. The smaller $\mathrm{d}c$, the closer to the envelope we get.
This means that the point on the envelope is at the intersection of $$ f(x,c)=0 $$ and $$ \frac{\partial}{\partial c}f(x,c)=0 $$
Example
The family of lines parametrized by $a$: $$ \frac{x}{1-a}+\frac{y}{a}=1 $$ Take the derivative with respect to $a$: $$ \frac{x}{(1-a)^2}-\frac{y}{a^2}=0 $$ Solve simultaneously $$ x=(1-a)^2\qquad y=a^2 $$
The family of lines is in black and the envelope is in green. The envelope follows the intersection of adjacent curves.