Is there a proof (apart from contradiction) to show that identity does not exist. I have example in which there is only option of contradiction to prove that identity does not exist.
Consider the binary operation : $a*b= 2a +b, \forall a,b \in \mathbb{Z}$.
 Suppose that $\mathbb{Z}$ has an identity $e$ w.r.t. $*$. Then, $e*1 = 2e + 1  = 1 \implies  2e = 0 \implies   e=0$.
But, $1*0 = 2 + 0 = 2 \ne 1$, so $0$ cannot be an identity.
 A: If you take as an axiom that identities are unique (call it proposition $A$), then there is a unique element $e$ such that $e*x=x$ for all $x\in \mathbb{Z}$ (call it porposition $B_{1}(e)$) and $x*e=x$ for all $x\in \mathbb{Z}$ (call it proposition $B_{2}(e)$).
In symbols: $A\Rightarrow (\exists e\in \mathbb{Z}(B_{1}(e) \wedge B_{2}(e)))$
Equivalently: $(\forall e\in \mathbb{Z}(¬B_{1}(e) \vee ¬ B_{2}(e)))\Rightarrow ¬A$
In other words, if you prove that for every element $e\in \mathbb{Z}$ such that $e*x=x$ for all $x\in \mathbb{Z}$ there exists an element $y\in \mathbb{Z}$ such that $y*e \neq y$, then you can deduce that an identity does not exist.
Warning: this is not a proof by contradiction, it is a proof by contrapositive. We used the implication $A\Rightarrow B$ in its equivalent form $¬ B \Rightarrow ¬ A$. But if I remember the logic course I took correctly, to prove this equivalence you are actually using the falsum deduction rule. So you are doing some sort of proof by contradiction implicitly.
If you are interested you can read this question.
A: $e \in \mathbb Z$ is identity if for every $a \in \mathbb Z,$ $$a*e=e*a=a.$$
To prove directly that $\mathbb Z$ doesn't have identity we need to show that no element of $\mathbb Z$ is identity. So we need to show that no element of $\mathbb Z$ satisfies the above equalities for every $a.$ 
That means we need to show that for every $x \in \mathbb Z,$ there exists $a \in \mathbb Z$ such that either $x *a \neq a$ or $a *x \neq a.$
Let $x \in \mathbb Z.$ 
If $x =0,$ then take $a=1.$ So $a*x=2 \neq a.$
If $x \neq 0,$ then take $a=x.$ So, $x*a=2x+a \neq a.$
