Terminology regarding elements of monoids In what follows, the symbols $a,b$ and $n$ implicitly range over $\mathbb{N} = \{0,1,2,\cdots\}.$
Are there names for the following properties that an element $x$ in a monoid may or may not possess? Both 1 and 2 are implied by 3. Conversely, 1 and 2 together imply 3.


*

*$x^n = 1 \rightarrow n=0$.

*$x^a = x^b \rightarrow x^{a-1} = x^{b-1}$ whenever $a,b > 0$.

*$x^a = x^b \rightarrow a=b$
Proof that 1 and 2 imply 3. Suppose $x^a = x^b$ and $a < b$, and assume $x$ has properties 1 and 2. Then using 2, we can show by induction that $1 = x^{b-a}$. Thus using 1, we have $b-a=0$. So $a=b$, a contradiction.
Remark. We can define that an element $x$ in a monoid is weakly cancellative iff whenever both $xa=xb$ and $ax=bx$ hold, we have $a=b$. Then every weakly cancellative element has property 2.
 A: I don't know if there is some particular term in the literature somewhere, but the following observation is certainly important: $x\in M$ (a monoid) has your property #3 if and only if $\langle x \rangle \cong \mathbb{N}$.
Rephrased: $x$ generates a monoid isomorphic to $\mathbb{N}$, the free monoid with one generator.
So one might conceivably refer to this in the following way: "$x$ is free," or "$x$ is a free element of $M$."
To say the same thing again, differently: if $M$ is any monoid and $x\in M$ is any element, then there is a unique monoid homomorphism $\varphi_x : \mathbb{N} \to M$ sending $1$ to $x$.  All of your properties have to do with this map $\varphi_x$.  1) says that it has no kernel, 3) says that it is injective.
A: If you exclude groups then 2 and 3 are equivalent.
Consider the monogenic semigroup $\langle x\rangle$ generated by $x$ (its structure is described in Wikipedia). It is either infinite (isomorphic to $\mathbb{N}$) or defined by a relation $x^{m+r}=x^r$ $\ (m,r>0)$.
Your monoid is either  $\langle x\rangle$ (if $\langle x\rangle$ has an identity element) or a semigroup with extra joined identity $\langle x\rangle\cup \{1\}$. In the first case the monoid is a group (since its idempotent contains in the kernel, which is an ideal). So we can consider the second case.
If $\langle x\rangle$ is infinite it yields 1-3 evidently. If it is finite then:
$3$ is not true since   $x^{m+r}=x^r$ don't imply $m+r=r$;
$2$ is not true since   $x^{m+r}=x^r$ don't imply  $x^{m+r-1}=x^{r-1}$
$1$ is true (rather meaningless).
Conclusion: 2 and/or 3 mean that $x$ is torsion-free; 1 means that is not a group.
