The story goes like this: A friend and I found this old exercise:
Let $G=\Bbb R-\{-1\}$ and $a*b:=a+b+ab$, is $(G,*)$ a group?
I say that $(G,*)$ is not a group because for any $a\in G$ follows that $0*a=0+a+0a=a+0+a0=a*0$ and also $1*a=1+a+1a=a+1+a1=a*1$, hence it has two identities. My friend says that it doesn't matter that it has two indentities since the definition we've got says that "there exists an element such that blah blah blah". But I say it does matters because then what happens with the inverse, something like this happens: $0=a*a^{-1}=1 \rightarrow 0=1$ (ugly!!!). Am I right? or his evil trolling mind is right?
Or in an (absurd) way is the definition not correct?
(There should be a tag called [settle-argument])
I'm of course joking
$1*a=2a+1\not\equiv a$ and so $1$ is not an identity. Instead, what you demonstrated is that $1$ is a non-trivial center, which all elements other than $0$ will be since this operation is abelian. It is not possible for a set to have two distinct left-right identities, since if $e_1$ and $e_2$ are identities, then $e_1*e_2=e_1$ and $e_1*e_2=e_2$ and so $e_1=e_2$.