Addendum 1: Note that I was talking here about the $\mathbb Z_\ell$ (and not $\mathbb Q_\ell$ Tate modules. You can divide up the classification of elliptic curves in two stages: First you see if the $V_\ell$ are isomorphic (and there it is enough to look at a single $\ell$). If they are, then the curves are isogenous. Then the second step is to look within an isogeny class and try to classify those curves.

The way I am talking about here goes directly to looking at the $T_\ell$ for all $\ell$. If they are non-isomorphic (for even a single $\ell$ then the curves are not isomorphic and if they are isomorphic for all $\ell$ they still may or may not be isomorphic, the difference between them is given by a rank $1$ locally free module over the endomorphism ring. In any case they are certainly isogenous. These can be seen a priori as if all $T_\ell$ are isomorphic so are all the $V_\ell$ but also a posteriori essentially because a rank $1$ locally free module becomes free of rank $1$ when tensored with $\mathbb Q$.

Of course the a posteriori argument is in some sense cheating because the way you show that the curves differ by a twist by a rank $1$ locally free module is to use the precise form of the Tate conjecture:$$\mathrm{Hom}(E_1,E_2)\bigotimes \mathbb Z_\ell = \mathrm{Hom}_{\mathcal G}(T_\ell(E_1),T_\ell(E_2))$$which for a single $\ell$ gives the isogeny.

Note also that the situation is similar (not by chance) to the case of CM-curves. If we look at CM-elliptic curves with a fixed endomorphism ring, then algebraically they can not be put into bijection with the elements of the class group of the endomorphism ring (though they can analytically), you have to fix one elliptic curve to get a bijection.

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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.

Addendum: 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 $\mathrm{Hom}(E_1,E_2)\bigotimes_{\mathrm{End}(E_1,E_1)}E_1$ \mathrm{Hom}(E_1,E_2)\bigotimes_{\mathrm{End}(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)$\mathrm{End}(E_1)$ 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.

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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.

Addendum: 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 $mathrm{Hom}(E_1,E_2)\bigotimes_{\mathrm{End}(E_1,E_1)}E_1$ \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_1E_2$is determined by$\mathrm{Hom}(E_1,E_2)$and every rank$1\$ projective module appears in this way.