If all Tate modules (i.e., for all $\ell$) are isomorphic then they differ by
the twist by a locally free rank $1$ module over the endomorphism ring of one of
them. This is true for all abelian varieties but for elliptic curves we only
have two kinds of possibilities for the endomorphism ring; either $\mathbb Z$ or
an order in an imaginary quadratic field. In the first case there is only one
rank $1$ module so the curves are isomorphic. In the case of an order we get
that the numbe of twists is a class number.

<b>Addendum</b>: Concretely, we have that $\mathrm{Hom}(E_1,E_2)$ is a rank $1$ projective module over $\mathrm{End}(E_1,E_1)$ (under the assumption that the Tate modules are isomorphic) and then $E_2$ is isomorphic to $mathrm{Hom}(E_1,E_2)\bigotimes_{\mathrm{End}(E_1,E_1)}E_1$ (the tensor product is defined by presenting $\mathrm{Hom}(E_1,E_2)$ as the kernel of an idempotent $n\times n$-matrix with entries in $\mathrm{End}(E_1,E_1)$ and $E_2$ is the kernel of the same matrix acting on $E_1^n$. Hence, given $E_1$ $E_2$ is determined by $\mathrm{Hom}(E_1,E_2)$ and every rank $1$ projective module appears in this way.