Spectral theory in Hilbert spaces usually starts with the assumption that an unbounded operator is closed. This is a restriction applied to the class of all unbounded operators in a separable Hilbert space. Why is it needed? Can't one define the resolvent operator and the split of $\mathbb{C}$ into the resolvent and the spectral sets if the operator is unbounded and completely arbitrary?
I do not recall reading a justification of studying spectral theory only for closed operators.
If the operator $T$ is bijective with a bounded inverse, then its graph satisfies $$ \Gamma(T)=\{(x,Tx)\mid x\in D(T)\}=\{(T^{-1}y,y)\mid y\in H\}. $$ The RHS is (up to a flip of the coordinates) the graph of $T^{-1}$, which is closed. Thus $T$ must be closed.
Now assume that $\lambda\in \rho(S)$. If we apply the result from the first paragraph to $S-\lambda$, we obtain that $S-\lambda$ is closed. From this it is not hard to conclude that $S$ itself must be closed.
In other words, you can define the spectrum of an operator that is not closed, but it will be all of $\mathbb{C}$. That is why spectral theory is not very interesting if the operator is not closed.