Assume that in this regression Y=β0+β1x+ϵ, where ϵ follows a Poisson distribution. Using OLS, estimate β0,β1 and cov(β1,β0).
I am wondering does the distribution of errors changes the β0 and β1 OLS estimators? If we used the normal procedure of deriving β1,β0, the result would be the same, but I think I am wrong.
The distribution of the error terms $\varepsilon_i$ do not change the point estimates of the $\beta_0$ and $\beta_1$ coefficients in OLS. You'll note that the usual derivation of $\hat\beta_0$ and $\hat\beta_1$ do not use the the distribution of the error terms at all. The distribution of $\varepsilon_i$ matters when you are trying to estimate the variances of $\hat\beta_0$ and $\hat\beta_1$ though.
The distribution of $\varepsilon$ will also affect $Cov(\hat\beta_0, \hat\beta_1)$. Once you substitute the estimate $\hat\beta_0 = \bar Y - \hat\beta_1 \bar X$ into $Cov(\hat\beta_0, \hat\beta_1)$, this eventually simplifies to:
$$ Cov(\hat\beta_0, \hat\beta_1) = Cov(\bar Y, \hat\beta_1) - \bar X \cdot Var(\hat\beta_1)$$
And note that the value of $Var(\hat\beta_1)$ depends on $Var(Y_i) = Var(\varepsilon_i)$, so the distribution of $\varepsilon$ is relevant here.