Suppose that $T:D(T)\subseteq X\to Y$ is a potentially unbounded linear operator, and recall that the multivalued part of $T$ is defined as the set $\{y\in Y:T(0)=y\}$.
If $T$ is a 'genuine' linear operator, or well-defined as a linear operator, one trys to show that $\{y\in Y:T(0)=y\}=\{0\}$ only. That is, that the multivalued part of $T$ is just zero. The task here is to show the equality of these two sets, that is, to show (i) $\{y\in Y:T(0)=y\}\subseteq\{0\}$ and (ii) $\{y\in Y:T(0)=y\}\supseteq\{0\}$.
I believe that (i) is shown by working with the image of $0$ under $T$; that is, we take the operator $T$ as given, and show that $T(0)$ coincides with $0$. This is really the 'hard' part, usually. As for showing the other 'easy' direction, I was wondering if it would be sufficient to use the fact that $\ker(T):=\{x\in D(T):Tx=0\}$ is always nonempty from the linearity of $T$? Because then, one has that $0\in\ker(T)\implies T(0)=0\implies0\in\{y\in Y:T(0)=y\}$, that is, $\{0\}\subseteq\{y\in Y:T(0)=y\}$.
Is this sufficient for the other direction?