Given a Hilbert or Banach space $\mathcal{H}$.
Consider two closed operators $S:\mathcal{D}(S)\to\mathcal{H}$ and $T:\mathcal{D}(T)\to\mathcal{H}$.
Suppose they're isometric on a common core $\mathcal{D}$: $$\|Sz\|=\|Tz\|\quad z\in\mathcal{D}$$ Then they already had a common domain $\mathcal{D}(S)=\mathcal{D}(T)$.
Especially they're isometric: $$\|Sx\|=\|Tx\|\quad x\in\mathcal{D}(S)=\mathcal{D}(T)$$
That is a simple limiting argument but how do I prove that they had a common domain?
I now looked up the definition of a core of an operator and it could be that the answer actually depends on the exact definition.
In Wikipedia, a core (of a closed operator) is defined as a subset $D$ of the domain of $A$ such that $A$ is the closure of $A|D$.
If we replace this by requiring that $D$ be a subspace, I can prove your claim.
By symmetry, it suffices to show $D(S) \subset D(T)$.
Let $x\in D(S)$ be arbitrary. There is a sequence $(x_n)_n$ in $D$ with $(x_n, Sx_n) \to (x,Sx)$. Hence,
$$ \Vert Tx_n - Tx_m \Vert = \Vert T (x_n - x_m)\Vert = \Vert Sx_n - Sx_m \Vert \to 0 $$
for $n,m \to \infty$. Here I used $x_n - x_m \in D$, because $D$ is a subspace.
As we are working on Banach spaces, this yields $(x_n, Tx_n) \to (x,y)$ for some $y$. But $T$ is assumed closed, which in particular yields $x \in D(T)$.