I apologize for in advance for making just a few superificial remarks. These are:
The question is not uninteresting. Just because Sha doesn't appear in the definition of L easily, there's no reason one shouldn't ask about manifestations more fundamental than the usual one.
An approach might be to think about the p-adic L-function rather than the complex one. I'm far from an expert on this subject, but the algebraic L-function is supposed to be a characteristic element of a dual Selmer group over some large extension of the ground field. The Selmer group (over the ground field) of course does break up into cosets indexed by Sha. Perhaps one could examine carefully the papers of Rubin, where various versions of the Iwasawa main conjectures are proved for CM elliptic curves.]
Added, 8 July:
This old question came back to me today and I realized that I had forgotten to make one rather obvious remark. However, I still won't answer the original question.
You see, instead of the $L$-function of an elliptic curve $E$, we can consider the zeta function $\zeta({\bf E},s)$ of a regular minimal model ${\bf E}$ of $E$, which, in any case, is the better analogue of the Dedekind zeta function. One definition of this zeta function is given the product
$$\zeta({\bf E},s)=\prod_{x\in {\bf E}_0} (1-N(x)^{-s})^{-1},$$
where ${\bf E}_0$ denotes the set of closed points of ${\bf E}$ and $N(x)$ counts the number of elements in the residue field at $x$. It is not hard to check the expression
$$\zeta({\bf E},s)=L(E,s)/\zeta(s)\zeta(s-1)$$
in terms of the usual $L$-function and the Riemann zeta function.
The product expansion, which converges on a half-plane, can also be written as a Dirichlet series
$$\zeta({\bf E},s)=\sum_{D}N(D)^{-s},$$
where $D$ now runs over the effective zero cycles on ${\bf E}$. This way, you see the decomposition
$$ \zeta({\bf E},s)=\sum_{c\in CH_0({\bf E})}\zeta_c({\bf E},s), $$
in a manner entirely analogous to the Dedekind zeta. Here, $CH_0({\bf E})$ denotes the rational equivalence classes of zero cycles, and we now have the partial zetas
$$\zeta_c({\bf E},s)=\sum_{D\in c}N(D)^{-s}.$$
It is a fact that $CH_0({\bf E})$ is finite. I forget alas to whom this is due, although the extension to arbitrary schemes of finite type over $\mathbb{Z}$ can be found in the papers of Kato and Saito.
It's not entirely unreasonable to ask at this point if the group $CH_0({\bf E})$ is related to $Sha (E)$. At least, this formulation seems to give the original question some additional structure.
Added, 31, July, 2010:
This question came back yet again when I realized two errors, which I'll correct explicitly since such things can be really confusing to students. The expression for the zeta function in terms of $L$-functions above should be inverted:
$$\zeta({\bf E},s)=\zeta(s)\zeta(s-1)/L(E,s).$$
The second error is slightly more subtle and likely to cause even more confusion if left uncorrected. For this precise equality, ${\bf E}$ needs to be the Weierstrass minimal model, rather than the regular minimal model. I hope I've got it right now.
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 $\mathrm{Hom}(E_1,E_2)\bigotimes_{\mathrm{End}(E_1)}E_1$ (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 $\mathrm{End}(E_1)$ 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.
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.
Best Answer
Let me first give you a heuristic "reason", why the regulator in the class number formula looks different from the regulator in the Birch and Swinnerton-Dyer conjecture. It is often more convenient (and more canonical) to combine the regulator and the torsion term in the Birch and Swinnerton-Dyer conjecture: one chooses a free subgroup $A$ of the Mordell-Weil group of $E(K)$ with finite index and looks at the quantity $R(A)/[E(K):A]^2$, where $R(A)$ is the absolute value of the determinant of the Néron-Tate height pairing on a basis of $A$. As you can easily check, the square in the denominator insures that this quantity is independent of the choice of $A$. Now, in the class number formula, the torsion term is replaced by the number of roots of unity in $K$ and that term is not squared. So for such a canonical formulation to be possible, it is reasonable to expect that not the regulator of the number field should be defined through a symmetric pairing on the units, but its square. Then you could make the same definition as in the elliptic curves case, and it would be independent of the choice of finite index subgroup.
Now, that we have established this, there are several ways to bring the two situations closer together. You could do the naïve thing: take the matrix $M=(\log|u_i|_{v_j})$, where $u_i$ runs over a basis of the free part of the units (or more generally $S$-units for any set of places $S$ which includes the Archimedean ones), and where $v_j$ runs over all but one Archimedean place (or more generally all but one place in $S$), v_0, say. The absolute values have to be suitably normalised (see e.g. Tate's book on Stark's conjecture). Now take the symmetric matrix $MM^{tr}$ and define this to be the matrix of a new pairing on the units. In other words, you would define your symmetric pairing as $$ (u_1,u_2) = \sum_{v\in S\backslash\{v_0\}} \log|u_1|_v\log|u_2|_v. $$ Then, it's clear that you have a symmetric pairing and that the determinant of that pairing with respect to any basis on the free part of the units is $R(K)^2$. As I mentioned above, the square was expected.
Depending on what you want to do, this pairing might not be the best one to consider. For example if now $F/K$ is a finite extension and you consider the analogous pairing on the $S$-units of $F$ and restrict it to $K$, then it's not clear how to compare it to the pairing on $K$. In the elliptic curves case by contrast, the former is $[F:K]$ times the latter. To fix this, we can make the following definition: $$ (u_1,u_2) = \sum_{v\in S} \frac{1}{e_vf_v}\log|u_1|_v\log|u_2|_v, $$ where $e$ and $f$ are the absolute ramification index and residue field degree respectively. With this definition, the compatibility upon restriction to subfields is the same as in the elliptic curves case. On the other hand, the relationship with the actual regulator is slightly less obvious. It is however quite easy to show (and I have done it in http://arxiv.org/abs/0904.2416, Lemma 2.12, in case you are interested in the details) that the determinant of this pairing is given by $$ \frac{\sum e_vf_v}{\prod e_vf_v}R(K)^2, $$ with both the sum and the product again ranging over the places in $S$.
So I guess, the moral is that one shouldn't seek analogies between the BSD-formula and the class number formula but rather between the BSD and the square of the class number formula (note that also sha, which is supposed to be the elliptic curve analogue of the class number, has square order whenever it is finite). There is also a corresponding heuristic on the analytic side.