I think there's an interesting parallel in multivariable calculus that may somehow justify the terminology (very loosely).
If $\mathbf{G} = P\mathbf{i} + Q\mathbf{j}$, and $C$ is the boundary of a simply-connected region $D$ in the plane, then Green's Theorem gives:
$ \int_C \mathbf{G} \cdot \mathrm d\mathbf{r} = \int_C P \mathrm dx + Q \mathrm dy = \iint_D \left( \frac{\partial}{\partial x} Q - \frac{\partial}{\partial y}P \right)\mathrm dA$.
Then we also have an identity for $\mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}$ involving the $\mathbf{curl}$ of the field:
$ \int_C \mathbf{F} \cdot \mathrm d\mathbf{r} = \int_C P \mathrm dx + Q \mathrm dy + R \mathrm dz = \iint_D \mathbf{curl}(\mathbf{F}) \cdot \mathbf{k} \mathrm dA$.
Observe that $\dfrac{\partial}{\partial x}Q - \dfrac{\partial}{\partial y}P$ can be regarded as the determinant of $\begin{bmatrix}\frac{\partial}{\partial x} & \frac{\partial}{\partial y} \\ P & Q \end{bmatrix}$, while $\mathbf{curl}(\mathbf{F})$ is the determinant of
$\begin{bmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R\end{bmatrix}$,
which is conveniently notated $\mathbf{\nabla} \times \mathbf{F}$. This gives some (very stretched) credence to the "2-dimensional" cross-product, as defined in your question above, as we could then write:
$\mathbf{\nabla} \times \mathbf{G} = \begin{vmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \\ P & Q \end{vmatrix} = \frac{\partial}{\partial x} Q - \frac{\partial}{\partial y}P $.
Ok, I realize that in the 3-dimensional case, there was also the dot product with $\mathbf{k}$, but like I said, this is a loose connection.