$(a) = R$ if and only if a is a unit. Let $⟨R;+,−,0,·,1⟩$ be a commutative ring. For $a \in R$, define $(a)$ $:= \{a · r | r \in R \}.$ How can i prove that $(a) = R$ if and only if a is a unit. 
So if there exist  $a' \in R$ such that $a*a' = 1$ we have $ \forall r \in R $ $ (a*a')*r = r$ hence $(a) = R$ , right ?
If $(a) = R$  than  $ \forall r \in R $ there exists $r' \in R $ such that $ a*r' = r$ but than ...?
Is the first part correct and how can i formulate the second part?

The first part is correct. 
For the second one, your initial thought needn't be true: as an example, consider $R=\mathbb{Z}_p$ and the ideal $(-1)$. Then, even though $(-1) = \mathbb{Z}_p$, in general $(-1)r \neq r$. 
The only implication is that $\forall r \in R \quad\exists r'$ such that $ar=r'$.From this, putting $r'=1$, you see that $a$ must be a unit.
