Question 3)
In notation to be understood as matrix-vector, assume that the correct specification is
$$y = X\beta + \gamma y_{-1}+ e$$
(where $X$ contains the constant and the $X_1$ variable and $e$ is white noise, and $E(e\mid X) =0$), but you specify and estimate instead
$$y = X\beta + u$$
i.e. without including the LAD, and so in reality $u =\gamma y_{-1}+ e$.
Then OLS estimation will give
$$\hat \beta = (X'X)^{-1}X'y = (X'X)^{-1}X'(X\beta + \gamma y_{-1}+ e) $$
$$= \beta + (X'X)^{-1}X'y_{-1}\gamma +(X'X)^{-1}X'e$$
The expected value of the estimator is
$$E(\hat \beta) = \beta + E\Big[(X'X)^{-1}X'y_{-1}\gamma\Big] +E\Big[(X'X)^{-1}X'e\Big]$$
and using the law of iterated expectations
$$E(\hat \beta) = \beta + E\Big(E\Big[(X'X)^{-1}X'y_{-1}\gamma\Big]\mid X\Big) +E\Big(E\Big[(X'X)^{-1}X'e\Big]\mid X\Big)$$
$$= \beta + E\Big((X'X)^{-1}X'E\Big[y_{-1}\gamma\mid X\Big]\Big) +E\Big((X'X)^{-1}X'E\Big[e\mid X\Big]\Big)$$
$$=\beta + E\Big((X'X)^{-1}X'E\Big[y_{-1}\gamma\mid X\Big]\Big) + 0 $$
the last term being zero per our assumptions. But $E\Big[y_{-1}\gamma\mid X\Big] \neq 0$, because $X$ contains all the regressors (from all time periods), and so there is correlation with the LAD vector. Therefore $E(\hat \beta) \neq \beta$. In other words, ignoring the lag dependent variable will not make the estimator unbiased, as long as $\gamma \neq 0$, i.e. as long as the LAD does belong to the regression.
Question 1)
Assume now that you specify correctly, and denote $Z$ the matrix containing also the LAD.
Here (using the same steps as before)
$$\hat \beta = \beta + (Z'Z)^{-1}Z'e$$
and
$$E(\hat \beta) = \beta + E\Big((Z'Z)^{-1}Z'E\Big[e\mid Z\Big]\Big)$$
But is $e$ (the vector) independent of $Z$? No, because $Z$ contains the LAD from all time periods bar the most recent, while $e$ contains the errors from all time periods bar the first. So even if $e$ is not serially correlated, it is correlated with the vector $y_{-1}$.
So indeed, the last term is not zero and $$E(\hat \beta) \neq \beta$$ the OLS estimator is biased.
But the OLS estimator will be consistent if indeed the inclusion of the LAD eliminates serial correlation, because (using the properties of the plim operator)
$$\operatorname{plim}\hat \beta = \beta + \operatorname{plim}\left(\frac 1{n-1} Z'Z\right)^{-1}\cdot \operatorname{plim}\left(\frac 1{n-1}Z'e\right)$$
Part of the standard assumptions (and rather "easily" satisfied), is that the first plim of the product converges to something finite. The second plim written explicitly is (and using the stationarity assumption to invoke the LLN)
$$\operatorname{plim}\left(\frac 1{n-1}Z'\mathbf e\right) = \left[\begin{matrix}
\operatorname{plim}\frac 1{n-1}\sum_{i=2}^ne_i \\
\operatorname{plim}\frac 1{n-1}\sum_{i=2}^nx_{i}e_i \\
\operatorname{plim}\frac 1{n-1}\sum_{i=2}^ny_{i-1}e_i \\
\end{matrix}\right] \rightarrow\left[\begin{matrix}
E(e_i) \\
E(x_{i}e_i) \\
E(y_{i-1}e_i) \\
\end{matrix}\right]\; \forall i$$
$E(e\mid X) = 0 \Rightarrow E(e_i) = 0$, and also that $E(x_{i}e_i)=0$, for all $i$.
Finally, IF serial correlation has been removed, then $E(y_{i-1}e_i) =0$ also. So this plim goes to zero and therefore
$$\operatorname{plim}\hat \beta = \beta$$
i.e. the OLS estimator is indeed consistent in this case. So the "summary" is correct.
Question 2)
The full sentence from Wooldridge is
"It is also valid to use the SC-robust standard errors in models with lagged dependent variables assuming, of course, that there is good reason for allowing serial correlation in such models".
meaning, when we have good reasons to believe that the inclusion of lagged dependent variables does not fully remove autocorrelation. And it seems we got ourselves a Catch-22: if serial correlation (SC) has been removed, why use SC-robust std errors? And if serial correlation has not been removed, our OLS estimator will be inconsistent, so in such a case is it meaningful/useful/appropriate to use asymptotic inference? Well, it appears that if we do suspect that SC still exists, it is better to try to do something about it, regardless. But your comment has merit, and I would suggest to contact Wooldridge directly on the matter, in order to get an authoritative answer.
That would indeed be surprising, and although @Aksakal is right that without serial correlation, the estimates should not differ too much, it would still be a fluke if they were exactly equal (and I presume you have run your code more than once).
One possible explanation (hence at best a partial answer) could be that your bandwidth (I am not sure how they are precisely chosen) is such that your HAC estimates boil down to HC estimates, which, in turn, boil down to the classical ones when only regressing on a constant:
The robust variance matrix is
$\hat V= n(X'X)^{-1}(X'\hat\Omega X)(X'X)^{-1}$
with $\hat\Omega$ the covariance matrix of the residuals. Here, $X$ is just a column of 1s. Hence, $(X'X)^{-1}=1/n$. Now, if $\hat\Omega$ is a diagonal matrix (possibly because the weights are such that only the off-diagonal elements receive zero weight), we obtain
$$X'\hat\Omega X=\sum_{i=1}^n\hat u_i^2$$
such that
$\hat V= n\frac{1}{n}\sum_{i=1}^n\hat u_i^2\frac{1}{n}=\frac{1}{n}\sum_{i=1}^n\hat u_i^2$
Upon inserting the degrees of freedom correction $n/(n-1)$ used in the default OLS formula, we get
$$\frac{1}{n-1}\sum_{i=1}^n\hat u_i^2.$$ In R (I do not have access to MATLAB):
library(sandwich)
n <- 10
y <- rnorm(n,1)
reg <- lm(y~1)
vcov(reg)
vcovHC(reg,type="HC0")*n/(n-1)
Best Answer
Newey-West standard errors are asymptotically consistent, meaning that the estimated variance-covariance matrix should converge to the true one.
Why do you suspect that you have non-zero off-diagonal elements of your true variance covariance matrix? Newey-West usually assumes that the rows of your model matrix are ordered from earliest to latest observations and the size of the correlation is inversely related to the number of rows apart the observations are. Is this consistent with the correlation structure that you envision for your data?