operation on set proof

Question

Consider the operation ⊥ defined by placing, for every $x,y\in Z$
$x⊥y=x+|y|$, Check Associativity and Commutativity. Is there a Identity element in $Z$?
My proof:
Associativity
$x⊥(y⊥z)=(x⊥y)⊥z$
$x⊥(y⊥z)=x+|y+|z||$
$(x⊥y)⊥z=(x+|y|)+|z|=x+|y|+|z|\not=x+|y+|z||$
Commutativity
$x⊥y=y⊥x$
$x⊥y=x+|y|$
$y⊥x=y+|x|\not=x+|y|$
Identity element
$x⊥e=e⊥x=x$
$x⊥e=x+|e|=x\Rightarrow e=0?$
$e⊥x=0+|x|\not=x$
so no condition has been met?