I just want to see if I understood this. Since the geometric product is associative and so on we can write for two multivectors A and B given by
$A= \alpha_{0}+\alpha_{1}e_{1}+\alpha_{2}e_{2}+\alpha_{3}e_{1}\wedge e_{2}$
$B= \beta_{0}+ \beta_{1}e_{1}+\beta_{2}e_{2}+\beta_{3}e_{1}\wedge e_{2}$
the geometric product multiplication as
$AB=M=\mu_{0}+\mu_{1}e_{1}+\mu_{2}e_{2}+\mu_{3}e_{1}e_{2}$
Where for example $\mu_{0}=\alpha_{0}\beta_{0}+\alpha_{1}\beta_{1}+\alpha_{2}\beta_{2}-\alpha_{3}\beta_{3}$ and so on.
Now let's take an example with simple vectors with numbers like $a = (e_{1}+2e_{2}), a_{1}=(2e_{1}+3e_{2}), a_{2}=(2e_{1}+0e_{2})$
So $aa_{1}= 8- e_1\wedge e_{2}$ Now what if I multiply this with $a_{2}$?
$(8-e_{1}\wedge e_{2})(2e_{1})$? Is it just $16e_{1}-(e_{1}\wedge e_{2})(2e_{1})$ ? My logic behind this is that one can symbolically (GP) multiply multivectors and then opens up his list with like 20 different products (vector/bivector, bivector/bivector, trivector/bivector ect.) or something and then evaluates each of the products? Is this right?
EDIT: I just saw lmsteffan's comment. It seems that you probably did mean for $e_3$ to be $e_2$, in which case your calculation is correct. Oops. Oh, well.
I don't know how you got the result you did from the product of your vectors, but it's wrong. The thing that should really help you out is that the geometric product is distributive. That along with associativity, and a few other important things like that orthogonal vectors commute under this product, let's us figure out how to calculate it.
So let's consider your vectors $a = (e_{1}+2e_{2}), a_{1}=(2e_{1}+3e_{3})$. Their product is then $$aa_1 = (e_{1}+2e_{2})(2e_{1}+3e_{3})=(e_1)(2e_1)+(e_1)(3e_3)+(2e_2)(2e_1)+(2e_2)(3e_3)=2(e_1e_1)+3(e_1e_3)+4(e_2e_1)+6(e_2e_3) = 2(e_1 \cdot e_1)+3(e_1\wedge e_3)-4(e_1 \wedge e_2)+6(e_2 \wedge e_3)=2-4(e_1 \wedge e_2)+3(e_1\wedge e_3)+6(e_2 \wedge e_3)$$This is the answer, but let's go further. $$=(2+0+0)+e_1\wedge(-4e_2+3e_3) + (2e_2)\wedge (3 e_3)=(e_{1}+2e_{2})\cdot (2e_{1}+3e_{3}) + (e_{1}+2e_{2})\wedge (2e_{1}+3e_{3})$$ Notice that this last equation says that $aa_1 = a\cdot a_1 + a\wedge a_1$. This is called the Fundamental Identity of Geometric Algebra (for vectors). Now you should be able to see how to multiply this by $a_2$.
Note that I've assumed that your $e_i \bot e_j$ for $i\ne j$ and that $e_i \cdot e_i = 1$ for all $i$. The product gets a little more interesting if that isn't necessarily the case.