How to prove the existence of the identity element of an binary operator?

Question

In order to explain what I'm asking, let's consider the following binary operation:

The binary operation $*$ on $\mathbb{R}$ give by $x*y = x+y - 7$ for all $x,y$ $\in \mathbb{R}.$

In here it is pretty clear that the identity element exists and it is $7$, but in order to prove that the binary operation has the identity element $7$, first we have to prove the existence of an identity element than find what it is.

So, how can we prove that the existance of the identity element ?

Note: I actually asked a similar question before, but in that case the binary operation that I gave didn't have an identity element, so, as you can see from the answer, we directly proved with the method of contradiction.Therefore, instead of asking a new question, I'm editing my old question.

Answer 1

You guessed that the number $7$ acts as identity for the operation $*$. Then you checked that indeed $x*7=7*x=x$ for all $x$. Therewith you have a full proof that an identity element exists, and that $7$ is this special element.

Answer 2

Remark: the binary operation for the old question was $x*y = 3(x+y)$.

Suppose on the contrary that identity exists and let's call it $e$.

$$0*e = 3(0+e)= 0$$

$$3e=0$$

and hence $$e=0$$

However, $$1*e = 3(1+0)=1$$

and we obtain $$3=1$$ which is a contradiction.

Edit in response to the new question : $\forall x \in Q$, $x + 0 = x$ and $0+x= x$. Hence $0$ is the additive identity.

If you are willing to accept $0$ to be the additive identity for the integer and $\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$.

$\frac{a}{b}+\frac{0}{1}=\frac{a(1)+b(0)}{b(1)}=\frac{a}{b}$