Let E and F be two elliptic curves. For convenience, suppose that they do not have CM. For each p, there is a Galois representation

rho: = rho`_`

E + rho`_`

F :G_Q ---> GL`_`

2(F`_`

p) x GL`_`

2(F`_`

p)

given by the action of Galois on the p-torsion of E and F. For sufficiently large p, the Galois groups Q(E[p]) and Q(F[p]) are SL_2(F`_`

p)-extensions of Q(zeta`_`

p). If p is big enough, PSL`_`

2(F`_`

p) is simple. Thus, either:

1). These extensions are disjoint, and im(rho) = kernel of (det(x)/det(y)).

2). These extensions intersect in a have a common PSL_2(F`_`

p) extension of Q(zeta`_`

p), and hence a PGL`_`

2(F`_`

p) extension of Q. This implies that their projective representations are the same. This implies that they are equal up to twisting by a character. Since their determinants are also the same, this character is either trivial or quadratic.

If E[p] = F[p] infinitely often then E is isogenous to F, using Falting's proof of the Tate conjecture.

If Case 1) occurs, one can count that the density of pairs of elements (sigma,tau) in im(rho) such that Trace(sigma) = Trace(rho) is ~1/p. By Cebotarev, this implies that the number of primes l such that a(E,L) = a(F,L) mod p has density at most ~1/p. Hence, if 1) occurs infinitely many times, then a(E,L) = a(F,L) for set of density zero.

If E[p] is a non-trivial quadratic twist of F[p] for infinitely many primes, looking at the ramification of E and F it is easy to see that E[p] = F[p] tensor chi some fixed quadratic character chi for infinitely many p. Yet then using Faltings again, E is isogenous to the twist of F by chi.

PS: This is a pretty standard argument.

PPS: The argument requires the existence of Galois representations (EDIT: with big image!) attached to E (and F), which do not exist in general (EDIT: for example, weight one forms, see Toby's answer). The argument basically works for any pair of classical modular forms e,f of weight at least 2.