Noether theorem with semigroup of symmetry instead of group

As lurscher says, semigroups are not too interesting in physics. They're not really "symmetries". An important physics example of a semigroup in physics is the (misleadingly called) "Renormalization Group" that allows us to derive effective laws for long distances from the short-distance ones, but this "integrating out" or "flowing" is irreversible - which is exactly why the RG operations deserve to constitute a "semigroup" rather than a "group".

However, more importantly, you are totally confused about the case of Noether's observable called "energy".

Energy is associated with the time-translational-invariance and it is a standard symmetry, expressed by a group and not "just a semigroup". This symmetry says that the laws of physics don't change if the events occur $t$ seconds later. Also, they didn't change if the processes occurred $t$ minutes earlier. The time translations go in both directions, they can be both positive and negative, and they're always symmetries.

This has nothing to do with the irreversibility of the processes or the asymmetries associated with the arrow of time.

However, you are actually wrong even about the time reversal. Even the time reversal symmetry, which is a symmetry of the microscopic laws of physics (if CP holds; or at least CPT is such a symmetry universally) is a symmetry, and it is again expressed by a $Z_2$ group. There is no non-group semigroup anywhere in those spacetime-related operations!

Back to why "semigroups" are not symmetries

Quite generally, objects such as the action cannot be symmetric under semigroups that are not groups. The example of the Renormalization Group clearly shows what happens if the would-be symmetry transformations were irreversible. Irreversibility always means that some information is getting lost, so the contributions to the action from the degrees of freedom whose information is getting lost are inevitably being suppressed, so that action cannot stay constant. If one "flows" to long distances via the Renormalization Group flows, he actually "integrates out" the shortest-distance degrees of freedom which means that they cannot be excited in the new effective theory. So the operation cannot provide us with a one-to-one map for all states which, as explained in the previous sentences, is necessary for a symmetry, and symmetries therefore have to groups and not just semigroups.

Here is an argument why "a Noether Theorem with Lie monoid symmetry" essentially wouldn't produce any new conservation laws. Noether's (first) Theorem is really not about Lie groups but only about Lie algebras, i.e., one just needs $n$ infinitesimal symmetries to deduce $n$ conservation laws. If one is only interested in getting the $n$ conservation laws one by one (and not so much interested in the fact that the $n$ conservation laws together form a representation of the Lie algebra), then one may focus on a $1$-dimensional Abelian subgroup of symmetry. The corresponding Lie subalgebra then becomes just $u(1)\cong\mathbb{R}$. Now returning to the question, one may, of course, artificially truncate a Lie group into a Lie monoid, say, if $q$ is a cyclic variable for a Lagrangian $L$, then artificially declare that the symmetry monoid is $q \to q + a$ for only non-negative translations $a\geq 0$, while artificially denying all negative $a<0$. On the other hand, one needs at least access "from one side" because Noether's Theorem is about continuous symmetry. But in practice, one can then always extend, at least infinitesimally, to "the other side" as well, and then one is back to a standard $u(1)$ Lie algebra and a standard Noether Theorem.

The main problem with semi-groups is that they are not in general invertible, so it doesn't apply well to the notion of a symmetry, which is something that behaves the same after performing an operation that transforms the states into each other