I have proved that it is for $a>0$ by constructing six parametrizations that are diffeomorphisms. When $a=0$, these parametrizations lose their diffeomorphic character as the derivative fails to exist at the origin.
However, the definition of a manifold is that ${\bf there \ exists}$ a local diffeomorphism from an open neighborhood of the set to $\mathbb{R}^k$ everywhere.
So just because the specific parametrizations that I constructed fails doesn't disprove that the set isn't a manifold. Obviously at the vertex of a cone, the object cannot map smoothly to $\mathbb{R}^2$; but how would I formally prove this using the definitions of manifolds, diffeomorphisms, parametrizations, smoothness, etc?
If it were a manifold, the point $P$ (the origin) would have a basis neighbourhood on the induced topology (which is an intersection of a ball with the cone) homeomorphic to an open connected set of $\mathbb{R}^2$. However, taking away the point $P$ from such a basis neighbourhood would leave a not-connected set, whereas taking away the image of the point $P$ from the chart codomain will leave a connected set.
If you are only asking about the upper part of the cone, then it indeed admits a structure as a smooth manifold, since there is a global homeomorphism to $\mathbb{R}^2$ (any topological manifold with one single chart is obviously a smooth manifold). Just not one which makes it a submanifold of $\mathbb{R}^3$.