Anti-dual numbers. Based on similar principles you can build an algebraic system taking any symmetric (against both x and y axes) and crossing x axis at 1 and -1 curve as a unit circle.
Using lemniscate you get anti-hyperbolic numbers. Using diagonal square you get numbers with taxicab/manhattan metric.
Let's take a Cartesian plane equipped also with polar coordinates.
Consider a contour around the origin, defined in polar coordinates as $r=r(\phi)$. The contour is symmetric against the $x$ and $y$ axis-es and crosses points $(1,0)$ and $(-1,0)$. The contour defines the set of points to which we assign magnitude $1$.
Now, an arbitrary point, corresponding to a 2-dimensional number on the plane $z=(a,b)$ is characterized by angle $\alpha(z)=\arctan(b/a)$, magnitude $M(z)=\frac{\sqrt{a^2+b^2}}{r(\alpha(z))}$ and argument $\operatorname{arg}(z)=\int_0^{\alpha(z)} r(\phi)^2 d\phi$.
The addition of numbers is defined element-wise as $(a_1,b_1)+(a_2,b_2)=(a_1+a_2,b_1+b_2)$.
The multiplication is defined in such a way that the arguments are added and magnitudes are multiplied: $\operatorname{arg}(uv)=\operatorname{arg}(u)+\operatorname{arg}(v)$ and $M(uv)=M(u)M(v)$. This uniquely gives us a point which corresponds to the product. Multiplication defined this way is commutative.
Given these definitions, the usual complex numbers correspond to $r(\phi)=1$, split-complex numbers correspond to $r(\phi)=\frac{1}{\sqrt{\cos ^2(\phi)-\sin ^2(\phi)}}$ and dual numbers correspond to $r(\phi)=\frac1{|\cos \phi|}$.
Now, the most promising shapes are the reciprocals of the shapes of split-complex and dual numbers, which are lemniscates:
$r(\phi)=\sqrt{\cos ^2(\phi)-\sin ^2(\phi)}$

In this system $M(z)=\frac{a^2+b^2}{\sqrt{2 a^2-\left(a^2+b^2\right)}}$ and $\operatorname{arg} z=\frac{a b}{a^2 + b^2}$
(this formula is valid only for the first quarter, otherwise a piecewise definition is needed).
and
$r(\phi)=|\cos(\phi)|$

In this system the magnitude is $M(z)=\frac{a^2 + b^2}a$ and argument is $\operatorname{arg} z=\frac{1}{4}\left(\frac{a b}{a^2+b^2}+\arctan\left(b/a\right)\right)$ (this formula is valid only for the first quarter, otherwise a piecewise definition is needed).
As well as hexagonally symmetrical "hyperbola"
$r(\phi)=\frac{1}{\sqrt{\cos ^2\left(\frac{3 \phi }{2}\right)-\sin ^2\left(\frac{3 \phi }{2}\right)}}$

It seems that the first two have infinitely large elements, for instance, in the second one it is $\omega=(0,1)$. The third one has zero divisors and (seemingly) idempotents.