The basis itself is a function of $\theta$.
Equations 1 and 2 are fine, but there is a cross term in the dot product you're interested in.
To work this out you should first convince yourself that $\begin{eqnarray} \partial_r \hat e_r &=& 0, \\ \partial_r \hat e_\theta &=& 0, \\ \partial_\theta \hat e_r &=& \hat e_\theta \\ \partial_\theta \hat e_\theta &=& -\hat e_r. \end{eqnarray}$ The cross term comes from $(\frac{1}{r} \hat e_\theta)\cdot [(\partial_\theta \hat e_r)\partial_r] = \frac{1}{r}\partial_r$.
Let's look at this in more detail. First notice that $\hat{e}_{r}\partial_{r} + \frac{1}{r}\hat{e}_{\theta}\partial_{\theta}$ is the del operator in polar coordinates. Thus, you are trying to find the Laplacian in polar coordinates.
FOIL out the expression $\left(\hat{e}_{r}\partial_{r} + \frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right) \cdot \left(\hat{e}_{r}\partial_{r} + \frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right).$ You will find four terms, $\begin{eqnarray} \left(\hat{e}_{r}\partial_{r}\right) \cdot \left(\hat{e}_{r}\partial_{r}\right) &=& \partial_r^2 \\ \left(\hat{e}_{r}\partial_{r}\right) \cdot \left( \frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right) &=& 0 \\ \left(\frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right) \cdot \left(\hat{e}_{r}\partial_{r}\right) &=& \frac{1}{r} \partial_r \\ \left(\frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right) \cdot \left(\frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right) &=& \frac{1}{r^2}\partial_\theta^2. \end{eqnarray}$ The first and fourth terms are the ones you have claimed. But there is a subtlety. For example, $\begin{eqnarray} \left(\frac{1}{r}\hat{e}_{\theta} \partial_{\theta}\right) \cdot \left(\frac{1}{r}\hat{e}_{\theta} \partial_{\theta}\right) &=& \left(\frac{1}{r}\hat{e}_{\theta}\right)\cdot \left( \frac{1}{r} \hat{e}_{\theta} \partial_{\theta}^2 + \frac{1}{r} (\partial_\theta \hat e_\theta) \partial_\theta \right) \\ &=& \left(\frac{1}{r}\hat{e}_{\theta}\right)\cdot \left( \frac{1}{r} \hat{e}_{\theta} \partial_{\theta}^2 - \frac{1}{r} \hat e_r \partial_\theta \right) \\ &=& \frac{1}{r^2} \partial_\theta^2, \end{eqnarray}$ since $\hat e_\theta\cdot \hat e_r = 0$. Notice that $\partial_\theta (\hat{e}_{\theta} \partial_{\theta}) = \hat e_\theta \partial_\theta^2 + (\partial_\theta \hat e_\theta)\partial_\theta$ by product rule of differentiation.
Being similarly careful with the other terms you will find the claimed result, $\left(\hat{e}_{r}\partial_{r} + \frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right)^2 = \partial_r^2 + \frac{1}{r} \partial_r + \frac{1}{r^2} \partial_\theta^2.$