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

\[\rho: = \rho_E + \rho_F : G_{\mathbb{Q}} \to \mathrm{GL}_2(\mathbb{F}_p) \times \mathrm{GL}_2(\mathbb{F}_p)\]

given by the action of Galois on the $p$-torsion of $E$ and $F$. For sufficiently large $p$, the Galois groups $\mathbb{Q}(E[p])$ and $\mathbb{Q}(F[p])$ are $\mathrm{SL}_2(\mathbb{F}_p)$-extensions of $\mathbb{Q}(\zeta_p)$. If $p$ is big enough, $\mathrm{PSL}_2(\mathbb{F}_p)$ is simple. Thus, either:

1). These extensions are disjoint, and $\mathrm{im}(\rho) = \ker(\det(x)/\det(y))$.

2). These extensions intersect in a have a common $\mathrm{PSL}_2(\mathbb{F}_p)$ extension of $\mathbb{Q}(\zeta_p)$, and hence a $\mathrm{PGL}_2(\mathbb{F}_p)$ extension of $\mathbb{Q}$. 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 $\mathrm{im}(\rho)$ such that $\mathrm{Tr}(\sigma) = \mathrm{Tr}(\rho)$ is $\sim 1/p$. By Cebotarev, this implies that the number of primes $\ell$ such that $a(E,\ell) = a(F,\ell) \pmod{p}$ has density at most $\sim 1/p$. Hence, if 1) occurs infinitely many times, then $a(E,\ell) = a(F,\ell)$ 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 $E[p] = F[p] \otimes \chi$ for 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 $2$.