I am trying to define a group structure on the twisted cubic $C$ as follows
$$\begin{array}{ccc} C\times C&\longrightarrow C\\ ([p_0,..,p3],[q_0,...,q3])&\longmapsto & \begin{cases} [p0q0,...,p3q3] \text{ if } \ p_0q_0\neq 0, p_3q_3\neq 0\\ [1,...,1] \end{cases} \end{array}$$
the objective is to see that this map is regular, it is obvious at $C\times C\setminus Z(p_0q_0,p_3q_3)$ but I don't see if the map is regular at $([1,0,0,0],[0,0,0,1])$, maybe because it's not?
Is it possible to define another group structure?
There is no such group structure. Indeed, the twisted cubic is isomorphic to $\Bbb P^1 \cong S^2$ has Euler characteristic $2$, and an algebraic group which is projective has Euler characteristic zero (I fact I believe it is diffeomorphic to a torus).
Edit : let me add some more justification. First, the Poincaré-Hopf theorem say that for $X$ compact one has $\chi(X) = \sum_z i_z(v)$ where $v$ is a generic vector field and $i_z(v)$ is the index of $v$ at $z$. If $X$ is also a group, then multiplication by a non-zero element induces a non-zero vector field everywhere, i.e $\chi(X) = 0$. Secondly, consider the map $f : G \to \mathfrak gl(\mathfrak g), g \mapsto (x \mapsto Ad(g)(x))$. This is an algebraic morphism from a projective variety to an affine variety, so it should be constant ! In particular, $Ad(0)(x) = 0$ so it follows that $f$ is identically zero, i.e $G$ is abelian. It is a classic result that compact connected abelian algebraic groups are torus.