I want to find the eigenvalue of the differential operator $D(f)=f'=λf$. By solving the differential equation $f'=λf$ I get the eigenfunction ${e^{\lambda t}}$ which means $D(e^{\lambda t})=\lambda e^{\lambda t}=λf$.
What I understand by this is that ${e^{\lambda t}}$ is the eigenvector but what I'm not sure if the eigenvalue can be $\lambda$ itself.
You should read it as "let $\lambda \in \mathbb{C}$, let us try to find an eigenfunction of $D$ with eigenvalue $\lambda$". Then you do some work and find $e^{\lambda t}$ is such an eigenfunction. So any $\lambda \in \mathbb{C}$ is an eigenvalue. (Switch $\mathbb{C}$ for $\mathbb{R}$ if you are just working over $\mathbb{R}$.)