When you say $c = ab$ is a bivector, you're taking $a \perp b$ implicitly. The geometric product will in general have both terms: $ab = a \cdot b + a \wedge b$, remember? A scalar and a bivector.
It is, admittedly, pretty hard to visualize what that is geometrically. Adding vectors to vectors or bivectors to bivectors is sensible; adding vectors and bivectors strains thinking (and I know; I tried to explain this at a talk and people repeatedly insisted it was nonsense, but how is adding real and imaginary numbers any more sensible?).
Beyond that, it is possible to think about the generalized wedge and dot products separately and to imagine a geometric interpretation. The wedge product of a vector and a bivector produces the trivector with magnitude of the parallelepiped, as you know. The dot product produces the vector in the plane that is orthogonal to the first vector.
You can think of the geometric product, then, as describing a decomposition: given a bivector $B$ and a vector $v$, $v \wedge B = v_\perp B$ extracts the part of the vector orthogonal to the surface and builds a volume from that. $v \cdot B = v_\parallel B$ extracts the part of the vector that lies in the plane and gives the orthogonal vector in the plane that could be used to reconstruct $B$ with.
You talk about sweeping vectors over each other to form the wedge product, and that's a sensible way of looking at things, but what image do we need to imagine the dot product of a vector and a bivector? I can picture taking the bivector like a flat disc of clay, putting my hands on it, and squeezing it along the direction of $v_\parallel$. That's the best I can conceive of it. That's what the dot product is, after all--a reduction operation, one that forms vectors from planes and planes from volumes.
Thinking of the dot product as "division" does seem like a sensible interpretation.
$\newcommand{\Cl}{\mathscr{Cl}(V)}$The problem with your definition of the wedge product in $\Cl$ is that it is an $n$-ary product of vectors. And so it's not clear to me how one would even discuss associativity of it (i.e. it doesn't allow one to distinguish between the product of a bivector and a vector vs the product of a vector and a bivector). So while $\Cl$ with the wedge product is an algebra -- it isn't an associative algebra like the exterior algebra is. Let me instead recommend a different definition based on grade projection (basically the same definition given in the works of Alan Macdonald):
To define the wedge product, we first define it in the case of single-grade elements. Let $A,B$ be an $r$-vector and $s$-vector, respectively. Then
$$A\wedge B := \langle AB\rangle_{r+s}$$
Now using this special case, we define the wedge product for all multivectors. Let $C,D$ be (potentially mixed grade) multivectors in $\Cl$. Then we define
$$C \wedge D := \sum_{j,k} \langle C\rangle_j \wedge \langle D\rangle_k$$
Proof that for all $A,B,C\in \Cl$, we have $A\wedge (B\wedge C) = (A\wedge B)\wedge C$:
Consider a $j$-vector $A$, $k$-vector $B$, and $l$-vector $C$. Then
$$(A\wedge B)\wedge C = \langle AB\rangle_{j+k}\wedge C = \langle \langle AB\rangle_{j+k}C\rangle_{j+k+l} = \langle (AB)C\rangle_{j+k+l}$$
where that last step holds because only the \langle AB\rangle_{j+k} part of $AB$ can contribute to $\langle (AB)C\rangle_{j+k+l}$.$^\dagger$.
Then $$(A\wedge B)\wedge C = \langle (AB)C\rangle_{j+k+l} = \langle A(BC)\rangle_{j+k+l} = A\wedge\langle BC\rangle_{k+l} = A\wedge(B\wedge C)$$
To finish the proof by allowing $A,B,C$ to be any multivectors in $\Cl$, just note that our definition of the wedge product is bilinear.$^\ddagger$$\ \ \ \ \square$
Now we'll show that with this definition, your proposition holds. Take $\wedge_C$ to be the wedge product in $\Cl$ and $\wedge_T$ to be the wedge product in $\Lambda V$ (i.e. the coproduct of the antisymmetric tensor powers of $V$).
Let $\{e_1, \dots, e_n\}$ be an orthogonal basis for $V$ wrt the symmetric bilinear form and construct bases for $\Cl$ and $\Lambda V$ in the obvious way from it. Let $\Gamma: \Cl \to \Lambda V$ be given by $$\Gamma(A_0 + A_1e_1 + A_2e_2 + \cdots + A_{12\cdots n}e_1\wedge_C e_2\wedge_C \cdots \wedge_C e_n) = A_0 + A_1e_1 + A_2e_2 + \cdots + A_{12\cdots n}e_1\wedge_T e_2\wedge_T \cdots \wedge_T e_n$$
By definition this is linear. It's also clearly invertible. To see that it's multiplicative, it suffices (due to linearity of $\Gamma$ and bilinearity of $\wedge_C$) to show that $\Gamma$ is multiplicative on blades. Let $A=A_{i_1\cdots i_p}e_{i_1}\wedge_C\cdots\wedge_C e_{i_p}$ be a $p$-blade and $B=B_{j_1\cdots j_r}e_{j_1}\wedge_C\cdots\wedge_C e_{j_r}$ an $r$-vector. Then $$\begin{align}\Gamma(A\wedge_C B) &= \Gamma(A_{i_1\cdots i_p}e_{i_1}\wedge_C\cdots\wedge_C e_{i_p}\wedge_CB_{j_1\cdots j_r}e_{j_1}\wedge_C\cdots\wedge_C e_{j_r}) \\ &= A_{i_1\cdots i_p}e_{i_1}\wedge_T\cdots\wedge_T e_{i_p}\wedge_T(B_{j_1\cdots j_r}e_{j_1})\wedge_T\cdots\wedge_T e_{j_r} \\ &= (A_{i_1\cdots i_p}e_{i_1}\wedge_T\cdots\wedge_T e_{i_p})\wedge_T(B_{j_1\cdots j_r}e_{j_1}\wedge_T\cdots\wedge_T e_{j_r}) \\ &=\Gamma(A_{i_1\cdots i_p}e_{i_1}\wedge_C\cdots\wedge_C e_{i_p})\wedge_T\Gamma(B_{j_1\cdots j_r}e_{j_1}\wedge_C\cdots\wedge_C e_{j_r}) \\ &= \Gamma(A)\wedge_T\Gamma(B)\end{align}$$
A linear, invertible, multiplicative map is by definition an isomorphism of algebras. Hence $\Cl$ is isomorphic to $\Lambda V$, as desired.
Another Interesting Tidbit:
$\Lambda V$ is actually (or at least can be) defined by a set of axioms like other types of vector spaces.
From nlab:
Suppose $V$ is a vector space over a field $K$. Then the exterior algebra $\Lambda V$ is generated by the elements of $V$ using these operations:
- addition and scalar multiplication
- an associative binary operation $\wedge$ called the exterior product or wedge product,
subject to these identities:
- the identities necessary for $\Lambda V$ to be an associative algebra
- the identity $v\wedge v = 0$ for all $v\in V$.
It is easily confirmed that $\Cl$ satisfies these properties under the definition of $\wedge$ given above (and thus why associativity is so important for $\wedge$). Thus, not only can we construct an isomorphism, but we can even consider $\Cl$ to just be $\Lambda V$ with extra structure (the Clifford product).
$\dagger:$ To prove this, note that it is known that if $\Bbb F$ is not characteristic $2$, then $V$ has an orthogonal basis wrt any symmetric bilinear form. Then you can construct a "standard" basis for $\Cl$ and decompose your multivectors wrt to it.
$\ddagger:$ By the linearity of the grade projection operator.
Best Answer
In the geometric algebra of $R^3$ the bivectors are ismorfic to the complex numbers since any basis bivector $e_{ij} = e_i \wedge e_j = e_i * e_j$ multiplied by itself is $e_{ij}*e_{ij} = -1$.
$(e_i*e_j) * (e_i*e_j) = -e_j*(e_i *e_i)*e_j = -e_j*e_j = -1$
The product involved is the geometric product. Geometric product is defined for two vectors $a$ and $b$ as:
$a b = a \cdot b + a \wedge b$
Conversely, it is easy to show that:
$a \wedge b = 0.5( a b - b a)$
It is also easy to show that the cross product of two vectors is:
$a \times b = (a \wedge b) * e_{321}$
The dual of any bivector $B$ is a vector $c$ which is normal to the plane defined by $B$ itself and is defined as $c = B * e_{321}$.
Since
$e_1 = e_{23} * e_{321}$
$e_2 = e_{31} * e_{321}$
$e_3 = e_{12} * e_{321}$
An even grade multivector of the form $\alpha + B$, where $\alpha$ is a scalar (0-vector) and $B$ is a bivector or $2-vector$, is called a rotor. Rotors are isomorphic to quaternions as you pointed out. A unit rotor represent a rotation in 3D space.
The conversion between basis bivectors and conplex numbers $i$, $j$, $k$ is as you described.