Due to abstract nonsense, I would say for the purposes of dynamics any semigroup action of a monoid can be considered to be a monoid action (indeed any semigroup action of a semigroup can be considered to be a monoid action of a monoid, by adding an extra element if necessary to be and act as identity).
Say $(M,e)$ is a semigroup with identity (i.e. monoid), $X$ is a set, $\alpha_\bullet:M\to \operatorname{End}_{\text{Set}}(X)$ is a family of self-maps of $X$ satisfying the semigroup property
$$\alpha_{ts}=\alpha_t\circ \alpha_s.$$
Like you said it is not necessarily the case that $\alpha_{e}=\operatorname{id}_X$. However the semigroup property still limits the possibilities for $\alpha_e$, in that:
$$\alpha_e\in\{f\in\operatorname{End}_{\text{Set}}(X)\,|\, \forall t\in M: \alpha_t\circ \alpha_e=\alpha_t=\alpha_e\circ \alpha_t\},$$
in particular $\alpha_e$ has to be idempotent:
$$\alpha_e\circ \alpha_e=\alpha_e.$$
For $f:X\to X$ an idempotent, define an equivalence relation $\sim_f$ on $X$:
$$x\sim_f y\iff x=y \quad\text{ or }\quad f(x)=y\quad\text{ or }\quad x=f(y) \quad\text{ or }\quad f(x)=f(y).$$
(Reflexivity and symmetry of this relation is straightforward; for transitivity one can draw a $4\times 4$ table whose columns are all possibilities for $x\sim_f y$ and whose rows are all possibilities for $y\sim_f z$; idempotence of $f$ is important here. Succinctly, $y\sim_f x \iff y\in f^{-1}(f(x))$, that is, $y$ is in the $f$-saturation of $x$.)
Let us denote by $\pi^f:X\twoheadrightarrow X/\sim_f$ the canonical projection to the associated quotient set. Since $f(x)=f(x)$, $f(x)\sim_f x$, so that $f/\sim_f = \operatorname{id}$, i.e.
In particular, for $f=\alpha_e$, $\underline{\alpha_e}: X/\sim_{\alpha_e}\to X/\sim_{\alpha_e}, [x]\mapsto [\alpha_e(x)]$ is the identity map. Finally for $t\in M$, $\alpha_t$ preserves $\sim_{\alpha_e}$, so that
$$\underline{\alpha_t}: X/\sim_{\alpha_e}\to X/\sim_{\alpha_e},\,\, [x]\mapsto [\alpha_t(x)]$$
is well-defined. It is straightforward that now we have a monoid action $\underline{\alpha_\bullet}: (M,e)\to \left(\operatorname{End}_{\text{Set}}(X/\sim_{\alpha_e}),\operatorname{id}_{X/\sim_{\alpha_e}}\right)$.
My estimation is that (possibly with some natural modifications) this argument can be extended to actions with structure (measurable, topological, smooth, ...). So it is benign to assume that any semigroup action of a monoid is in fact a monoid action.
Finally note that in dynamics literature monoid actions are at times extended without notice to group actions even (using some version of the natural extension/ inverse limit/ solenoid construction discussed at Proving a.e surjectivity of suggested factor map in Natural extension of Standard Borel dynamical system, Does any measure preserving system have an invertible extension?, Projective limit of spaces of probability measures is bijective to the space of probability measures on a projective limit., or the aforementioned book of Arnold, pp.547-548).
Best Answer
For $\oplus$: It's not enough to show that $A\oplus A = A$. The definition of the identity is:
An identity is such an element $e$ that for all $a\in\mathbb B$, it holds that $ae=ea=a$.
This means it is not enough to find an element for each $a$, you must fint one element for all $a$. Your $0$ is a good candidate, can you show that $a\oplus 0 =0\oplus a= a$ for all $a\in\mathbb N$?
In the same way, you didn't yet show that $\ominus$ forms a monoid. You must find some element $e$ for which $e\ominus a = a$. If you cannot find it, you have to prove it does not exist.