For any transformation of the fields, $$\varphi\to\varphi'=\varphi+\delta\varphi$$ the change in the Lagrangian can be written as $$\delta\mathcal L = \text{EoM} + \partial_\mu j^\mu\tag{1}$$where "EoM" represents the equations of motion (Euler-Lagrange equations) and all other terms can be written as a total derivative of some function $j^\mu$, which is a known function in terms of the Lagrangian.
I would like to distinguish the different realizations of transformations. Let's assume that the transformation (1) leaves the action invariant, $\delta S=0$.
$\delta\mathcal L=0$
EoM $=0$, "on-shell": Noether current is conserved, $\partial_\mu j^\mu=0$.
EoM $=\partial_\mu b^\mu\neq0$, "off-shell": modified Noether current $J^\mu = j^\mu+b^\mu$ is conserved, $\partial_\mu J^\mu=0$.
$\delta\mathcal L =\partial_\mu a^\mu \neq 0$, "quasi-symmetry"
EoM $=0$, "on-shell": modified Noether current $J^\mu = j^\mu-a^\mu$ is conserved, $\partial_\mu J^\mu=0$.
EoM $=\partial_\mu b^\mu\neq0$, "off-shell": modified Noether current $J^\mu = j^\mu-a^\mu+b^\mu$ is conserved, $\partial_\mu J^\mu=0$.
Is this listing correct?
What roles do the terms "on/off-shell" and "(quasi) symmetry" play in Noether's theorem?