Consistency of OLS when no intercept

Suppose I have a model $$y_i = \beta_0 + \beta_1 x_i + e_i$$ but instead I estimate $$y_i = \beta_1 x_i + u_i$$ using OLS. That is, I ignore the intercept. Working out the algebra, based on this post, we have

$$\hat{\beta}_1 = \beta_1 + \beta_0 \frac{\sum_{i=1}^n x_i}{\sum_{i=1}^n x_i^2} + \frac{\sum_{i=1}^n x_i u_i}{\sum_{i=1}^n x_i^2}$$

I know how to care care of the third term, but can you help me verify if the following is correct?

I can write

$$\frac{\sum_{i=1}^n x_i}{\sum_{i=1}^n x_i^2} = \frac{\frac{1}{n}\sum_{i=1}^n x_i}{\frac{1}{n} \sum_{i=1}^n x_i^2}$$

Using the continuous mapping theorem, the numerator converges in probability to $$E[x_i]$$ and the denominator $$E[x_i^2]$$.

Suppose I demean the data. Then $$E[x_i] = 0$$. Does that mean $$\hat{\beta}_1$$ is consistent even if I ignore the intercept as long as I demean $$x_i$$ but not $$y_i$$?

When you demeans $$x_i$$, you assure that the intercept is truly zero, by construction. So yes, in this case $$\hat{\beta}_1$$ is consistent, if it is consistent in the model without demeaning and including the intercept.