Identity is an equation which is true regardless of what values are substituted for any variables (if there are any variables at all).
The question is: $$\frac{x^2-a^2}{x-a}=x+a$$
Is this an identity?
This equation is not true if $x=a$.
And how it's called if not.
First of all, I suspect there's a typo: I think it should be $$"{x^2-a^2\over x-a}=x+a."$$ (The "$x^2$" is an "$x$" in the OP.) E.g., set $x=2, a=1$ - in the OP this would yield 1=3.
Second, as the comments have pointed out, even with that fix, it's false if $x=a$.
There's an important conceptual issue lurking here: there are some functions, which are not defined on all of $\mathbb{R}$, but which "ought" to be in the sense that there is a natural function, extending the given function, which is defined on all of $\mathbb{R}$. E.g., $f(x)={x^2-1\over x-1}$ agrees with $g(x)=x+1$ wherever the former is defined, and $g$ is defined everywhere; so in some sense we might think that $f$ actually is $g$. But it's not - even though $f$ can be naturally "completed," that would be a different function. The specific $f$ is not defined everywhere.
The term "identity" is vague without a specified domain. The above is not an identity for $x \in \mathbb{R}$ because it isn't true for $x = a$, but it is true for $x \in \mathbb{R} \setminus \{a\}$ because, by the definition, the statement is true for all $x$ in the domain.
This answer takes into account the typo Emilio Novati noted.
I suppose that your question is about $$ \dfrac{x^2-a^2}{x-a}=x+a $$
and this is an idenity for $x\ne a$ but is not an identity for $x=a$ because the Left side is not defined.
