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,
$$\left(\hat{e}_{r}\partial_{r} + \frac{1}{r}\hat{e}_{\theta}\partial_{\theta}\right)^2
= \partial_r^2 + \frac{1}{r} \partial_r + \frac{1}{r^2} \partial_\theta^2.$$