No, this is not always possible. Consider the complete graph $K_{7}$. If $K_{7}$ can be drawn so that all crossings meet at a single point, then there is a planar graph with $8$ vertices and at least $\binom{7}{2}=21$ edges. But this contradicts the fact that every planar graph $G$ has at most $3|V(G)|-6$ edges.
This argument actually shows that every graph $G$ with at least $3|V(G)|-2$ edges can never be drawn so that all crossings occur at the same point.