Regarding complex version of IEEE-754
It is true that no complex version of IEE754 exists, but the same design rules as for IEEE-754 can be applied for complex arithmetic. However, this is more subtle than it seems in the first place.
For example the direct cartesian implementation of the complex multiplication (i.e. $(a+ib)(c+id)$ does not work with $inf$. Consider for example $(\text{inf} + 0i)(\text{inf} + 0i)$. the correct result would obviously be $(\text{inf} + 0i)$. But the direct floating point evaluation will give $(\text{inf} +\text{NaN}i)$ which is not correct. So the implementation needs to take care of these corner cases to always generate a correct result.
Furthermore for special functions, it is even not always obvious what the correct behavior with respect to +0, -0, +inf, -inf and NaN should be. And for complex arithemtic you even have combinations of +0 + inf, +inf + nan in real and imaginary part and so on.
Some papers which may be useful are (Kahan is a very autoritative source on this topic)
You could of course decide to not have inf and/or NaN. But then your arithemtic is not algrebaically closed which means not all operations generate a result, but require some sort of exception handling.
Regarding the implementation
You are completely free how to implement the complex numbers, as real and imaginary pair or in polar form. Once the behavior of the complex arithmetic is defined (especially with repsect to the special float and complex float values, as discussed above) the implementation just has to give the correct results.
There are considerations such as speed, but that you know already. polar form is usually not used because addition is quite expensive and I believe there are also issues with accuracy. However, if this is not an issue and you only need to multiply and divide then polar form is the way to go I would say.
What most of the algebras that you mentioned have in common is that they are all examples of unital algebras. A unital algebra is an algebra with a multiplicative identity element, and that multiplicative identity element is denoted with the symbol $1,$ though this is a bit of an abuse of notation. We really should denote it as $\vec1.$
I do not know what the double numbers are, and the binarions are just another name for the complex numbers. The dual numbers are an alternative to the complex numbers. Consider the the vector space $\mathbb{R}^2$ with basis $\{e_0,e_1\}.$ To form an algebra, we want a bilinear product $\cdot,$ and so if we can determine the products of the basis vectors, bilinearity is sufficient to determine all the products in the vector space. Bilinearity, to put it simply, is a more technical way of addressing distributivity. We also want the algebra to be unital, i.e, that is, for $e_0\cdot{x}=x\cdot{e_0}=x.$ Hence we know $e_0\cdot{e_0}=e_0$ and $e_0\cdot{e_1}=e_1\cdot{e_0}=e_1.$ The remaining product to determine is $e_1\cdot{e_1}.$ We know that $e_1\cdot{e_1}=e_1^2=ae_0+be_1,$ so $e_1^2-be_1-ae_0=0.$ We can complete the square, so that $e_1^2-be_1+\frac{b^2}4e_0=\left(\frac{b^2}4+a\right)e_0=\left(e_1-\frac{b}2e_0\right)^2.$ Since $\frac{b^2}4+a$ are themselves real numbers, this limits what the possible products $e_1^2$ can be. The three different cases are $\frac{b^2}4+a\lt0,$ $\frac{b^2}4+a=0,$ and $\frac{b^2}4+a\gt0.$ These three cases determine the three possible unital algebras over $\mathbb{R}^2$ up to isomorphism. These cases give the complex numbers, the dual numbers, and the split-complex numbers, respectively.
Of course, nothing is limiting you to make your algebra unital. You can have other number systems, those systems are just not named and not particularly interesting as extensions of the real numbers. But you can study them and they are perfectly valid structures to work with.
Now, let us talk about quaternions and octonions. You should look into the Cayley-Dickson construction. The construction is a construction that systematically extends the real numbers to the complex numbers, and applying the construction again to the complex numbers produces the quaternions, and applying it to the quaternions produces the octonions, and so on. Thus, the reason these algebras have a dimension that is a power of $2$ is because of the way the Cayley-Dickson construction works.
But if you are willing to work with algebras not built from the Cayley-Dickson construction, then there is absolutely nothing wrong with using algebras of other dimensions. On YouTube, there is a very good video about the triplex numbers, a unital associative and commutative algebra over $\mathbb{R}^3.$ The basis is $\{1,i,j\}$ and one has $i\cdot{j}=j\cdot{i}=1,$ and $i^2=j$ and $j^2=i.$ This actually equivalent to making the basis $\{1,i,i^2\}$ and noting the rule that $i^3=1,$ thus uniquely determining the multiplication table.
Now, to answer your last question: why only consider $i^2$? Well, not all algebras focus on only $i^2,$ as shown above. However, if you do uniquely determine $i^2=r$ where $r\in\mathbb{R},$ then you cannot separately control what $i^3$ is, because then $i^3=ri,$ necessarily. You cannot have it any other way. In general, if $i^n=r$ with $r\in\mathbb{R}$ for some $n\in\mathbb{N},$ then $i^{m+n}$ is automatically determined for all $m\in\mathbb{Z},$ so you have no freedom to choose those individually. In the example you gave, you suggested, $i^2=0,$ which makes your algebra equivalent to the dual numbers. Thus, it is already impossible that $i^3=-1,$ because $i^3=i\cdot{i^2}=i0=0.$ You can have some number system where $i^3=-1$ and $i^2=0,$ but that number system will not be an algebra over anything, and the multiplication will not distribute over addition. This opens up a whole can of worms beyond the scope of your question.
If you want more on the subject, you should read on the subject of Clifford algebras, the theory that unifies all of these concepts together.
EDIT 1: I would also like to point out that if you want to get really creative with number systems, you are not limited to algebras over $\mathbb{R}.$ You can form algebras over any field. Why not try making $2$-dimensional algebras over the field with two elements $\mathbb{Z}_2$? What about algebras over $\mathbb{Q}$?
Best Answer
This structure cannot possibly be isomorphic to complex numbers; as　you say yourself, it is four-dimensional over $\mathbb R$ while the complex numbers are two dimensional.
What you're missing is that the bicomplex $j, k$ from the article you linked have $j^2 = 1$ and $k^2 = -1$, but your $j$ has $j^2 = -1$. So take your $i,j$ and define $$ i' = i,\quad j' = ij,\quad k' = j. $$ You will find using the basis $1,i',j',k'$ that your algebra is the bicomplex numbers exactly as described in the article.