Somewhere Colin M. Campbell noted:

> If $A$ is a semigroup defined as $$A=Sg(\pi)=\langle a_1,\cdots, a_d\mid u_1=v_1,\cdots,u_e=v_e\rangle $$ then the same generators with the same relations can also be interpreted as the presentation of the following group:
  $$A^*=Gp(\pi)=\langle a_1,\cdots, a_d\mid u_1=v_1,\cdots,u_e=v_e\rangle $$

So he consider a semigroup first and then its possible analogous group form. I am working on this ideas and asking if we can do converse direction? I mean can we start with a (for example) finite group such as $$D_8=\langle x,y\mid x^2=y^4=(xy)^2=1\rangle$$ and then construct a finite semigroup accordingly? I see, I can write $x^2=1$ as $x^3=x$ without any problems but I want to be sure about doing this job. Sorry if this question seems ridiculous for you. Thanks for your comments and your time.