Is np-complete an equivalence class?
The second definition isn't valid, since it's self-referential. Given the first definition, the second statement is an easy theorem that follows from the first definition.

Indeed, suppose we define NP-completeness using the first definition. Let's see that the second definition is true.

$\Longrightarrow$ If $L$ is NP-complete then for all $L'$ in NP, $L'$ reduces to $L$. In particular, SAT reduces to $L$. Since SAT is NP-complete by Cook's theorem, there exists an NP-complete problem $L'$ (namely SAT) that reduces to $L$.

$\Longleftarrow$ Suppose $L$ is in NP and some NP-complete problem $L'$ reduces to $L$. We will show that every problem $M$ in NP reduces to $L$, and so $L$ is NP-complete. Given a problem $M$ in NP, since $L'$ is NP-complete, $M$ reduces to $L'$. Since $L'$ reduces to $L$, $M$ also reduces to $L$, and we're done.

The proof used two properties: (1) there exists some NP-complete problem, (2) reducibility is transitive: if $A$ reduces to $B$ and $B$ reduces to $C$ then $A$ reduces to $C$.

Finally, any two NP-complete problems reduce to each other (via the definition), and in this sense they form an equivalence class of the reduction relation.