Let $R$ a ring without identity. So, $$(R\quad\text{does not have the identity})\Rightarrow (\nexists\;e\in R\;\text{such that}\;ea=a\;\forall a\in R)[\text{It's correct?}].$$
Now, if I found an element $e\in R$ and an element $a\in R$ such that $ea=a$, can I conclude that this contradicts the hypothesis according to which $R$ not have identity?
First of all, we write quantifiers before statements they are referring. The notation
$$\not\exists e\in R \text{ such that }ea=a\forall a\in R$$
is invalid, simply because it is unclear. We don't know if it means
$$\forall a\in R:\not\exists e\in R \text{ such that }ea=a$$ or $$\not\exists e\in R: \forall a\in R: \text{ such that }ea=a.$$
The two statements above are not equivalent, and are in fact very much different. The second statement is much stronger.
Second of all, the negation of the statement "$R$ has an identity" is the negation of the statement
$$\exists e\in R: \forall a\in R: ae=e$$
and the negation of this statement is the second of the two statements above.
Your argument that if you find some $e,a$ such that $ea=a$, is not enough to prove that $R$ has no identity, since even in rings without an identity, we can set $a=e=0$ and have $ea=a$.