By "left semigroup-joined-semigroup" I mean an algebraic structures $(S,\cdot,\*)$ such that both $\cdot,\*$ are associative, and the following property holds (see [this](http://jas.shahroodut.ac.ir/article_616.html) )
$$
x\*(y\cdot z)=x\*y\*z\;\; ; \;\; \forall x,y,z\in S?
$$


The right and two-sided cases and group-joined-semigroup are defined analogously.


If $\cdot,\*$ are two associative binary operations in $S\neq \emptyset$ such that $\cdot=\*$ or $x\*y=x$ (for all $x,y\in S$) then $(S,\cdot,\*)$ is a left semigroup-joined-semigroup, obviously. Now:


(A) Does anyone know some other **examples** for such algebraic structures?


(B) Any other **similar** algebraic structures?


(C) Any related **topics** and/or **references**?


**Example.** For every fixed real number $b\neq 0$, the structure $(\mathbb{R},+,+\_b)$ satisfies $x+\_by+\_bz=x+\_b(y+\_bz)=x+\_b(y+z)$, where 
$x+\_by:=x+y-b[\frac{x+y}{b}]$ (see [this](https://www.researchgate.net/publication/232263082_SOME_ALGEBRAIC_PROPERTIES_OF_b-PARTS_OF_REAL_NUMBERS) , [this](http://bims.iranjournals.ir/pdf_305_daa595cb8449dd1f6b4a0cb354aefa32.html) and [this](http://jas.shahroodut.ac.ir/article_616.html)). 



Regarding a new algebraic structure
One example is that $\cdot$ is ordinary multiplication of integers and $x\*y=(x\cdot y) \bmod b.$ One could as well let $\cdot$ be ordinary addition. That is essentially your example except that you are not requiring $b$ to be a positive integer.


More generally, Let $\sim$ be an equivalence relation on $S$ and $\cdot$ a compatible associative operation in that $x \cdot y \sim x' \cdot y'$ provided that $x \sim x'$ and $y \sim y'.$ Pick a system of class representatives. Then $x \sim y$ exactly if $[x]=[y]$ and compatibility condition is 


$[x\cdot y]=[[x] \cdot [y]]$ for all $x,y.$ 


Then defining $x\*y=[x \cdot y]$ satisfies your requirement since $$x\*(y\cdot z)=[x\cdot (y \cdot z)]=[x \cdot (y\*z)]=x\*(y\*z). $$ The middle step since $x \sim x$ and $y \cdot z \sim y\*z.$


Moreover $x\*y\*z=x\*(y\cdot z)=(x \cdot y) \*z.$ You just have to do $\*$ last. 


It *might* be that all examples are like this (but it isn't, see below). Given $\*$ and $\cdot$ meeting your condition, define $v \sim w$ to mean that $x\*v=x\*w$ for all $x.$ This is clearly an equivalence relation and $y\cdot z \sim y\*z$ for all $y,z$. 


It is not clear that $\cdot$ must be compatible , but it is provided $\cdot$ has a left identity $u$ with $u\cdot v=v$ for all $v.$ Then we can define class representatives by $[x]=u\*x $ and we have


$$ [x \cdot y]=u\*(x \cdot y)=u\*((u \cdot x) \cdot (u \cdot y))=u\*(u \cdot x) \* (u \cdot y)=u\*(u \* x) \* (u \* y) $$ And, $$[[x] \cdot[y]]=u\*((u\*x)\cdot (u\*y))= u\*((u\*x)\* (u\*y)).$$


**NOTES** Having $\sim$ be equality gives the example of $\*=\cdot.$ The example in the comments of $x\*y=a$ for some constant $a$ is the case that everything in $S$ is equivalent to everything else. I suppose also the case that $x\*y=x$ gives the case that everything in $S$ is equivalent to everything else. However here it is not the case that $x\*y$ can be described in term of $x\cdot y$


