Writing the $U(1)$ gauge transformation as coordinate transformation In quantum mechanics one can "always" write the way an operator acts on a wave function as a coordinate transformation. As an example we can look at unitary representation of the momentum operator
\begin{equation}
U=e^{\frac{i}{\hbar}p} \psi= \psi(x)+a\psi '(x)+\frac{a^2}{2}\psi '' (x)+...=\psi(x+a)
\end{equation}
with $p=\frac{\hbar}{i}\frac{d}{dx}$.

Is there a similar way to write the $U(1)$ gauge transformation $\psi(x) \rightarrow \psi'(x)=e^{-\frac{i}{\hbar}\Lambda(x)} \psi(x)$ as a transformation of the coordinates in the sense that $e^{-\frac{i}{\hbar}\Lambda(x)} \psi(x)=\psi(R(x))$ where $R$ is some kind of map or operator? (In the example with the momentum operator $R=x+a$)
 A: Not all unitary transformations can be written as coordinate transformations. Not sure where you heard this, but it is verifiably false.
By and large symmetries can be separated into two classes: internal symmetries and coordinate symmetries. The $U(1)$ transformation you're asking about is of the former type. Something like a Lorentz transformation would be of the latter type. For coordinate symmetries, what you say is true by definition and is actually realized by the action of such a $U$ on the coordinate basis, $|\boldsymbol{X}\rangle$.
What is true for internal symmetries is the following. Suppose $U(g)$ is a unitary operator forming a representation of some group $G$. That is, for any $g,h\in G$ we have $U(g)U(h)=U(gh)$. Then assuming this to be an honest symmetry of the system, $U(g)$ commutes with the Hamiltonian and hence our states can be labeled by the energy and a group index simultaneously. That is, our wavefunction will look something like $\psi_a$.
Now the corrected version of the assertion in the question would be the existence of a collection of matrices (not operators) $R_a^b(g)$ which form a representation of the group $G$, meaning $R^a_c(g)R^c_b=R^a_b(gh)$ (summation over repeated indices implied), such that
$$
U(g)\psi_a=R^b_a(g)\psi_b.
$$
