Noether's theorem states that if a system has a continuous symmetry, there is a quantity related to this symmetry, called the Noether charge, which is conserved.
It does not state anything on the fact that adding a constant term to a measurable quantity may or may not change the physical description of the system. Only some physical quantities in fact are defined up to a constant term (one can add a constant term without change the physics of the system).
These quantities are, for example, some forms of potential energy, angles and angular phases, but not entropy.
Entropy is not defined up to a constant term.
Adding a constant term to the entropy does change the physics of the system. For example, the 3rd law of thermodynamics states that the entropy of a perfect crystal (or an ideal gas) at zero temperature is zero. This has the very physical consequence that zero temperature can be reached only asymptotically.
Few clarifications of the Noether's theorem
Examples of continuous symmetries:
time invariance, translational invariance, rotational invariance.
Corresponding conserved charge:
energy, linear momentum, angular momentum.
Now, if a system is translational invariant (for example, an isolated and closed system), that means that any thermodynamical observable of the system is translational invariant, e.g., volume, temperature, energy, and entropy.
Note that in a closed system, any irreversible process breaks time invariance in a subtle way, since energy may still be a conserved quantity although entropy is not (it increases). This however does not constitute an exception of Noether's theorem.
Best Answer
Translational symmetry in the sense of the standard formulation of Noether theorems means that the Lagrangian is invariant under the action of the group of spatial translations. This is not the case in your example because $U$ does not admit such invariance.
However there is another, more physical, version of the idea of translational invariance for a physical system:
The class of solutions of the equation of motion is invariant under spatial displacements.
In other words, if $$x=x(t)$$ is a solution with initial conditions $$x(0)=x_0\:,\quad \frac{dx}{dt}|_{t=0}=\dot{x}_0\:,$$ the solution with initial conditions $$x(0)=x_0 + s\:,\quad \frac{dx}{dt}|_{t=0}=\dot{x}_0$$ must be $$x=x(t)+s$$ that is, the initial solution changed by the same initial given translation at each time $t$. This fact is by no means trivial.
This invariance requirement is valid for your example as you can directly check. However, since this invariance requirement is weaker than the one used in the standard version of Noether theorem, it does not imply that the momentum is conserved.
In Lagrangian formulation the two notions of invariance are not equivalent. The Noetherian one implies the second one but the converse implication is false. In Hamiltonian formulation they are equivalent provided we restrict ourselves to deal with canonical transformations.
The natural question however arises whether this weaker notion of invariance of your system implies the existence of a conserved quantity (diferent from the momentum). The answer is positive in our case. There is in fact another, weaker, version of Noether theorem stating that, if the Lagrangian is not invariant under the one-parameter ($\epsilon$) group of transformations $$x \to x_\epsilon\:, \quad \dot{x} \to \dot{x}_\epsilon = \frac{d}{dt}x_\epsilon$$ but, at first order in the parameter $\epsilon$, the transformed Lagrangian differs from the initial one just due to a total derivative $$\frac{d}{dt}f(t,x) = \frac{\partial f}{\partial x}\dot{x} + \frac{\partial f}{\partial t}$$ then there is a conserved quantity along the solution of the motion equations: $$I(t,x, \dot{x}) = \frac{\partial L}{\partial \dot{x}} \partial_\epsilon x_\epsilon|_{\epsilon=0} - f(t,x)\:.$$ The proof is a trivial generalization of the know classical one. In the considered case $$L(t,x, \dot{x}) = \frac{m}{2}\dot{x}^2 - ax-b\:.$$ Thus, since our group of transformations is $$x \to x+\epsilon\:, \quad \dot{x} \to \dot{x}\:,$$ we have $$\partial_{\epsilon}|_{\epsilon=0} L(t,x_\epsilon, \dot{x}_\epsilon)= -a = \frac{d}{dt} (-at)\:.$$ We conclude that there exists a conserved quantity. This is $$I(t,x, \dot{x}) = m \dot{x} + at\:.$$ A posteriori this is obvious from the equations of motion themselves, but it also arises form a weak symmetry of the Lagrangian.