I want to prove that $\left(e_1\wedge e_2\right)e_1 = -e_1 \left(e_1\wedge e_2\right)$.
I did this in two ways, but I didn't understand why the second way is wrong.
First way: $\left(e_1\wedge e_2\right)e_1 = -\left(e_2\wedge e_1\right)e_1 = -e_2\wedge \left(e_1e_1\right) = -e_2\left(e_1\cdot e_1 + e_1\wedge e_1\right) = -e_2$
And $-e_1 \left(e_1\wedge e_2\right) = -\left( e_1 e_1\right)e_2 = -e_2$
Second way: $\left(e_1\wedge e_2\right)e_1 = e_1\wedge\left(e_2 e_1\right) = e_1\wedge \left(e_2\cdot e_1 + e_2 \wedge e_1\right) = e_1\wedge e_2\wedge e_1 = 0$
You can't share the associative property between the wedge and geometric products like you're doing. The math just doesn't work that way. Here's how to figure out what $\left(e_1\wedge e_2\right)e_1$ and $-e_1 \left(e_1\wedge e_2\right)$ are (and whether they are equal).
First off I'm going to assume that $\{e_1, e_2\}$ is an orthonormal basis for $\Bbb R^2$. If that's not what you meant then just let me know.
$$\begin{align}(e_1\wedge e_2)e_1 &= (-e_2\wedge e_1)e_1 & \text{because } v\wedge w = -w\wedge v \\ &= (-e_2e_1)e_1 & \text{because } v\ \bot\ w \implies vw = v\wedge w \\ &= -e_2(e_1e_1) & \text{because } (uv)w = u(vw) \\ &= -e_2(1) & \text{because } v^2 = \|v\|^2 \\ &= -e_2\end{align}$$
Now let's look at $-e_1 \left(e_1\wedge e_2\right)$. I won't be writing down the reasons for each step here so make sure you understand them.
$$\begin{align}-e_1(e_1\wedge e_2) &= -e_1(e_1e_2) \\ &= -(e_1e_1)e_2 \\ &= -e_2 \\ &= (e_1\wedge e_2)e_1\ \ \ \ \square\end{align}$$
Note that this in no way proves that bivectors always anticommute with vectors. That just turns out to be what happens in this case. Though now that you've seen how this proof is done, you might find it enlightening (and fun) to find some general rules for when vectors and bivectors commute, anticommute, or neither.