To close this one:
When are the asymptotic variances of OLS and 2SLS equal?
A: Only when the "matrix of instruments" essentially contains exactly the original regressors, (or when the instruments predict perfectly the original regressors, which amounts to the same thing), as the OP himself concluded.
The variance of the IV estimator (generalized or not), when the OLS is also consistent, is never smaller then the variance of the OLS.
This is one of the main reasons why the "Hausman specification test" (or the Durbin-Wu-Hausman) is performed: because if the data show that the two estimators have the same probability limits (i.e if the null of the test is not rejected), then we should stick to OLS, which has a lower variance.
If the null is rejected (and we have other arguments that support the consistency of the IV estimator), then the OLS is deemed inconsistent, and then we go with the IV estimator, to buy consistency at the price of higher variance.
I make some additional assumptions and simplify notations, nothing of which should cause confusion. Suppose for simplicity that data is generated according to $Y = \beta X + \epsilon$, where all variables are $\mathbb R-$valued and $\epsilon$ has zero mean and variance $\sigma^2$. Assume $X$ has the necessary moments. We have $n$ independent copies of the pair $(Y, X)$; $x = [x_1, \dots, x_n]'$ and $y = [y_1,\dots, y_n]'$.
The OLS estimator of $\beta$ is
\begin{align}
\hat{\beta} &=y'x / x'x \\
&= (x\beta + e)'x/x'x \\
& = \beta + \frac{e'x}{x'x} \\
&= \beta + \frac{\frac{1}{n}\sum_i \epsilon_i x_i}{\frac{1}{n}\sum_i x_i^2}
\end{align}
where $e = [\epsilon_1, \dots, \epsilon_n]'$. The assertion that this approaches $\beta + {\rm Cov}(X, \epsilon) / {\rm Var}(X)$ as $n\to\infty$ is usually in the sense of convergence in probability. According to standard definitions, an estimator is consistent if it converges in probability to the true parameter value, i.e. in this case if $\hat{\beta} \to \beta$ in probability. Here, we had an extra term, in general non-zero, on the right hand side so the estimator is inconsistent.
On the other hand, we say that $\hat{\beta}$ is unbiased if $\mathbb E \hat{\beta} = \beta$. This statement has nothing to do with convergence. All the same, the expectation of the right hand side is again $\beta$ + some term which is not zero in general. Thus, $\hat{\beta}$ is also biased. This answers the first question.
Regarding the second question, notice that bias is usually regarded as a property of estimators, and $\beta$ is unknown so it's not an estimator. Therefore, it does not make sense to speak of the bias of $\beta$. If we are liberal in the usage of bias and let it apply to anything, we see that $\mathbb E \beta = \beta$ so it would be "unbiased".
Without further assumptions on the dependence between $X$ and $\epsilon$ there really isn't any way to tell what the answer to three is, I believe. I should say I did not have time to go through the calculations so I may be wrong.
In light of the discussion under the other answers: If one declares a new definition of consistency in terms of the variance approaching zero, I don't see how inconsistency under the assumption ${\rm Cov}(X, \epsilon) \neq 0$ can be either confirmed or disproved without more assumptions. It's also important to note that consistency in the standard definition says nothing about decreasing variance or decreasing bias. These are separate concepts and should not be mixed up.
Best Answer
The "bias" they are mentioning in appendix A of the article is not the omitted variables bias, but a bias due to possible endogeneity. (They mention it several times in the article, and specifically in page 11 of the PDF, in the bottom part).
The estimator formulas that are mentioned in the Appendix are the regular OLS estimator formulas for 2 variables, as can be found in page 8 here:
For $S_{i} = \alpha D_{i} + \beta P_{i} + u_{i}$:
$\widehat{\alpha}_{OLS} = \frac{Cov(D,S)Var(P) - Cov(D,P)Cov(P,S)}{Var(D)Var(P) - Cov(D,P)^{2}} = \alpha + \frac{Cov(D,u)Var(P) - Cov(D,P)Cov(P,u)}{Var(D)Var(P) - Cov(D,P)^{2}}$
$\widehat{\beta}_{OLS} = \frac{Cov(P,S)Var(D) - Cov(D,P)Cov(D,S)}{Var(D)Var(P) - Cov(D,P)^{2}} = \beta+ \frac{Cov(P,u)Var(D) - Cov(D,P)Cov(D,u)}{Var(D)Var(P) - Cov(D,P)^{2}}$
They stated the $\beta$ equals 0, and that's how they got the formulas in the article. Again - the possible bias that they are mentioning is because of a possible endogeneity, not because any ommision.