It seems like we need to distinguish linear span from conical hull.
Given a set $S$ of elements of a vector space $V$ (could be the polynomials, as in your example), the linear span of $S$ is defined to be
$$ \text{span}(S)=\left\{\left.\sum_{i=1}^k\lambda_iv_i\ \right|k\in\mathbb{N},v_i\in{S},\lambda_i\in\mathbb{R}\right\}. $$
In contrast, the conical hull of $S$ is defined to be:
$$ \text{cone}(S)=\left\{\left.\sum_{i=1}^k\alpha_iv_i\ \right|k\in\mathbb{N},v_i\in{S},\alpha_i\in\mathbb{R}_+\right\}. $$
The difference between these definitions is that the scalars in the conical hull must be non-negative.
So when you say:
Does there exist a set of basis vectors that spans the convex cone?
you're asking
Does there exist a set $S$ such that $\text{span}(S)=C$?
It's easy to see that $\text{span}(S)$ is a linear subspace of the vector space $V$. So the answer to the question above is yes if and only if $C$ is a linear subspace of $V$. A linear subspace is a convex cone, but there are lots of convex cones that aren't linear subspaces. So this probably isn't what you meant.
If instead, you ask
Does there exist a set $S$ such that $\text{cone}(S)=C$?
then the answer is yes, but the set $S$ may be infinite.
Based on what you said in the comments, it sounds like you are looking for a set $S$ such that $\text{cone}(S)=C$, and a vector $v$ which is orthogonal to every element of $S$. This will only be possible if $\text{span}(C)$ is a proper subspace of $V$. In this case, you can take $v$ to be a vector in the orthogonal complement of $\text{span}(C)$. Otherwise, $\text{span}(C)=V$, and there is no vector which is orthogonal to every vector in $S$.
Best Answer
A set of points is convex if the line segment connecting any two points of the set lies entirely in the set. You'd think of a shape as "concave" if you could connect two points by a line segment and part of that line segment was outside the set. So "convex" means "non concave".