Let me show that such a group, if it exists, can't be algebraic. Any complex algebraic group $G$ is an extension of an abelian variety $A$ by an affine group $H$, i.e. we have an exact sequence $1\to H\to G\to A\to 0$ (Chevalley's theorem). Suppose $G$ had every elliptic curve as a subgroup. All of those would project isomorphically to $A$, since none of them cam intersect $H$. So we may just as well restrict ourselves to the case $G=A$. But $A$ has only countably many Lie subgroups.
upd: here is a proof in the general case: A subgroup of $G$ which is an elliptic curve is contained in a maximal (compact) torus; moreover, since all tori are conjugate, the isomorphism classes of the elliptic curves they contain are the same. Now we use the above argument: every torus contains countably many (closed) Lie subrgoups.
