How to distinguish between different elements in a semigroup? I've recently started learning abstract algebra on my own and there is something that I'm struggling with at the moment.
Suppose we have a semigroup with two elements $a,b$ and a binary multiplication defined by $ab=ba=aa=bb=a$. Can we really say that $a$ and $b$ are different in the context of this semigroup? More generally, if we have a semigroup where $ax=bx$ and $ya=yb$ for all $x,y$, can we really say that $a$ and $b$ are different elements in the semigroup?
Edit: Looking at the answers given it seems like I've been a bit unclear about the thing I'm having issues with at the moment, so I'll try to rephrase and clarify some things. I do understand that the cancellation laws don't apply in general to semigroups, and as can be seen in the original question I don't have a problem understanding that the semigroup I gave actually is a semigroup.
The thing I have trouble with is the fact that in the example I gave, there seems to be the case that the only real difference between the objects $a$ and $b$ in the semigroup is their labels, rather than any inherent properties the elements have within the context of the semigroup. So my question might be better phrased the following way: if two elements $a$ and $b$ in a semigroup $s$ are such that $ax=bx$ and $ya=yb$ for all $x,y\in S$, is there a way to try to investigate whether the only real difference between $a$ and $b$ within the context of $S$ is their labeling? As someone pointed out in the comments to my original post, if they were different $a$ would solve the equation $xx=x$ whereas $b$ would not solve $xx=x$, making them truly distinct within the semigroup. This is the type of answer I was actually looking for in the beginning.
I hope that this edit clarifies some things. Maybe my way of thinking about this is misguided to begin with, but it feels unproductive to communicate past eachother.
 A: I am still unsure whether I understood your question correctly but the notion of a semigroup quasi-identity might help you. Indeed it seems you want to test whether a semigroup satisfies the quasi-identity
$$
 (ax = bx \wedge ya = yb) \to a = b
$$
A: Sure.
Let $S$ denote a set containing an element $z$ and define $st=z$ for all elements $s,t\in S$. 
Then the multiplication is evidently associative, so $S$ equipped with that multiplication is by definition a semigroup.
Further all products have the same value $z$.
There is no condition on $S$ (also if $S=\varnothing$ then vacuously all products are the same).
A: Yes. Consider the set $\{0,1\}$ with the operation $\star$ defined by $a\star b=0$.
This is clearly associative and $0\star 1=1\star 0=0\star 0 = 1\star 1=0$.
A: I will answer this question myself rather than restate it, as it was difficult for me to formulate the question in a clear manner when I first wrote it. At the time I was completely new to semigroups (and I'm still far from familiar enough to be an expert on the topic), and it was difficult for me to appreciate that the semigroup $S=(\{a,b\},\cdot)$ as defined in my question actually can describe a "real structure" with the given multiplication rule and where $a$ and $b$ were really distinct elements in a more "essential way" than just having different labels. Then I became a bit more familiar with representation theory, and this is how I finally managed to resolve my issues with the semigroup $S$.
We can represent $S$ faithfully using 3-by-3 matrices in the following way: let
$$a=\left(
\begin {matrix}
0 & 0 & 0  \\
0 & 0 & 0  \\
0 & 0 & 0  \\
\end {matrix}
\right)\quad\text{and}\quad b=\left(
\begin {matrix}
0 & 0 & 0  \\
1 & 0 & 0  \\
0 & 0 & 0  \\
\end {matrix}
\right),$$
and let $\cdot$ be the standard matrix product. It is intuitively clear that $a\neq b$, and that this difference isn't just about different labels. It is also straightforward to verify that, indeed,
$$aa=ab=ba=bb=a,$$
as it should be. So we can give a concrete realization of $S$ in this way, where the multiplication is not just something we made up on the spot to give an "easy" example of a group, but rather a multiplication that should be familiar to most people pondering these kinds of questions. It is worth noting that it is not possible to find a faithful matrix representation of $S$ using 2-by-2 matrices, but I leave it to the reader to figure out why that is ;)
