Electric displacement in a homogenous linear dielectric

You are not interpreting correctly what he says. Griffiths argues that if your entire space is filled with homogeneous dielectric material, i.e. $epsilon_{r}left(x,y,zright)=rm const.$, then you can ignore the dielectric material in your calculation of $boldsymbol{D}$. He states that you can calculate $boldsymbol{E}$ like in the vacuum case (and that's why he writes $boldsymbol{E}_{rm vac}$), and then your electric displacement field is just

$$boldsymbol{D}=epsilon_{0}boldsymbol{E}_{rm vac}$$

Now, if you want to find the actual electric field, you use the usual equation $boldsymbol{D}=epsilonboldsymbol{E}$. Why does this all thing work at all? Because of the following correspondence

$$boldsymbol{nabla}cdotboldsymbol{D}=rho_{rm f} :::{rm and}::: boldsymbol{nabla}timesboldsymbol{D}=0$$ $$updownarrow$$ $$boldsymbol{nabla}cdotboldsymbol{E}_{rm vac}=frac{rho_{rm f}}{epsilon_{0}} :::{rm and}::: boldsymbol{nabla}timesboldsymbol{E}_{rm vac}=0$$

In other words, $boldsymbol{D}$ and $epsilon_{0}boldsymbol{E}_{rm vac}$ satisfy the same equations and are thus the same.