Are you assuming that the algebra is associative? It seems you use $x(xy)=(xx)y=xy$ for the definition of this product, right? There are some other nice identities with a name. You could include a multiplication table for the basis $e_1=1,e_2=x,e_3=y,e_4=xy$.

Yes, I am talking about an associative algebra.

Four-dimensional unital associative algebras have been classified.

Amazing. Do you have a reference for this?

Yes, see the paper by Willem de Graaf, and the references therein.

The first two relations suggest "boolean algebra" and the last, commutator, relation suggests non-commutative, so maybe a skew Boolean algebra. Reference: Antonio Bucciarelli, Antonino Salibra.

@DietrichBurde this paper classifies nilpotent algebras. Is my algebra nilpotent? What is a nilpotent algebra?

Nilpotent algebra: en.wikipedia.org/wiki/Nilpotent_algebra

Ok, so it is not nilpotent as $x^n=x\ne 0$ for all $n$.

If $\mathrm{char}(k) \neq 2$, commutative and anticommutative together imply nilpotent of index 2.

@EricTowers I'm not understanding, this is niether commutative nor anticommutative.

When I look at a matrix representation of it, I do not see anything familiar.

You should specify that $xy\ne 0$, otherwise, $\mathbb{R}^2$ satisfies athe multiplication table.

A presentation never contains inequalities in its relations. The relations do not imply that $xy=0$, so $xy\neq 0$. But of course if one adds $xy=0$ as a relation (which means taking a quotient) then we can find a different algebra.

Also as the OP explained, since $x^n\neq 0$, the algebra is not nilpotent. Actually since $k\times k$ is a quotient which is not nilpotent, $A$ is also not nilpotent.

@CaptainLama $xy$ is a nilpotent.

That is irrelevant. In a nilpotent algebra every element is nilpotent.

I think, either $xy=0$ or $yx=0$

@DietrichBurde the paper you linked to classifies nilpotent algebras. This is not a nilpotent algebra. Did you actually have a reference for the associative, unital (not necessarily nilpotent) case?

@JoshuaTilley where from do you know it is not commutative? Or it is just a postulate?

@Anixx No, it is not postulated that it is noncommutative, it just is. You are misunderstanding the construction. If you would like to follow you need to grasp the notion of a presentation of an algebra.

@JoshuaTilley the only one non-commutative associative 4-dimensional algebra with nilpotents and idempotents are $2\times2$ matrices.

@Anixx This sure looks like a 4-d construction with a proper ideal, nilpotents, and idempotents. What, in your view, isn't correct?

@rschwieb let's see. There should be a matrix representation if this is correct, as for any finite-dimensional associative algebra.

Yes, I have also references for the classification of all associative algebras in dimension $4$, but the nilpotent ones are the most difficult part of it. Read the references in the introduction!

@Anixx Yeah, i wrote one up, and didnt' see any issues, nor anything familiar.

@Anixx This is a bit complicated, but take $R=T_2(F)$, and its radical $J$, then consider the triangular matrix ring $\begin{bmatrix}R&J \\ 0&R\end{bmatrix}$ and the subring whose matrices have equal entries. That seems 4-dimensional, noncommutative, having idempotents and nilpotents, and nontrivial ideals...

@rschwieb Okay, you are saying, it is split-complex plus two dual units.

Or maybe, not. Seems non-commutative.

@rschwieb anyway, your construction does not fit the description, because either xy=0 or yx=0 I think, your construction is isomorphic to the 2x2 matrices.

@Anixx It's obviously noncommutative. It contains a copy of the upper triangular matrices. Of course it does not fit the description: I was just giving it as a counterexample to your proposal. It was not an intended description of the one in the post. As I said, I do not see any interesting way to re-describe the ring in the post.

Well, it is not isomorphic to the 2x2 matrices but does not match the description in the post either.

@rschwieb ah, I see. Thanks for the clarification.