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Lastly, if you find yourself wondering about sufficient conditions for a monoid to be BF, then there might be just the thing for you in another thread: Among many others, free monoids and free abelian monoids are BF-monoids (this is obvious), and so is the multiplicative monoid of non-zero elements of a Noetherian integral domain (this is less obvious). Moreover, if $H$ is a unit-cancellative BF-monoid ("unit-cancellative" means that $xy=x$ or $yx=x$, for some $x, y \in H$, implies $y \in H^\times$) and $M$ is a submonoid of $H$ with $M^\times = M \cap H^\times$, then $M$ is a BF-monoid too (Theorem 2.22(iv) + Corollary 2.23 in the above preprint). In point of fact, all the examples in the OP are a special case of these.

Lastly, if you find yourself wondering about sufficient conditions for a monoid to be BF, then there might be just the thing for you in another thread: Among many others, free monoids and free abelian monoids are BF-monoids (this is obvious), and so is the multiplicative monoid of non-zero elements of a Noetherian integral domain (this is less obvious). Moreover, if $H$ is a unit-cancellative BF-monoid ("unit-cancellative" means that $xy=x$ or $yx=x$, for some $x, y \in H$, implies $y \in H^\times$) and $M$ is a submonoid of $H$ with $M^\times = M \cap H^\times$, then $M$ is a BF-monoid too (Theorem 2.22(iv) + Corollary 2.23 in the above preprint). In point of fact, some of the examples mentioned in the OP are a special case of these.

With this in mind, assume $H$ is a reduced BF-monoid, so that $\mathsf L_H(x)$ is finite and non-empty for all $x \in H$. Accordingly, pick $\bar x \in H \setminus \{1_H\}$, and let $n_{\bar x}$ be the maximum of $\mathsf L_H(\bar x)$. If $m$ be an integer $\ge 1 + n_x$ and $\bar x = x_1 \cdots x_m$ for some $x_1, \ldots, x_m \in H$, then $$ \mathsf L_H(x_1) + \cdots + \mathsf L_H(x_m) \subseteq \mathsf L_H(\bar x), $$ and hence $$ \max \mathsf L_H(x_1) + \cdots + \max \mathsf L_H(x_m) \le \max \mathsf L_H(\bar x) = n_{\bar x}, $$ which is possible only if $\mathsf L_H(x_i) = \{0\}$, and hence $x_i = 1_H$, for some $i \in [\![1,m]\!]$. []

With this in mind, assume $H$ is a reduced BF-monoid, so that $\mathsf L_H(x)$ is finite and non-empty for all $x \in H$. Accordingly, pick $\bar x \in H \setminus \{1_H\}$, and let $\rho(\bar x)$ be the maximum of $\mathsf L_H(\bar x)$. If $m$ is an integer $\ge 1 + \rho(\bar x)$ and $\bar x = x_1 \cdots x_m$ for some $x_1, \ldots, x_m \in H$, then $$ \mathsf L_H(x_1) + \cdots + \mathsf L_H(x_m) \subseteq \mathsf L_H(\bar x), $$ and hence $$ \max \mathsf L_H(x_1) + \cdots + \max \mathsf L_H(x_m) \le \max \mathsf L_H(\bar x) = \rho(\bar x), $$ which is possible only if $\mathsf L_H(x_i) = \{0\}$, and hence $x_i = 1_H$, for some $i \in [\![1,m]\!]$. []

Edit. In the comments to this answer, Mark Sapir writes, "This does not seem to answer the question. The OP likes idempotents and does not like units." So, let me try to clarify why this is an answer to the question, though not the only one possible (I'll continue with the same notations used in the above).

Given $x \in H \setminus \{1_H\}$, we denote by $\mathsf L_H(x)$ the set of all $k \in \mathbf N^+$ for which there exist atoms $a_1, \ldots, a_k \in \mathbf N^+$ such that $x = a_1 \cdots a_k$. Moreover, we take $\mathsf L_H(1_H) := \{0\}$. It can be proved (this is not for free) that ${\sf L}_H(u) = \emptyset$ for every $u \in H^\times \setminus \{1_H\}$. In addition, it is straightforward that $$ \mathsf L_H(x) + \mathsf L_H(y) \subseteq \mathsf L_H(xy),\ \text{ for all }x, y \in H. $$ Lastly, $H$ is a BF-monoid iff $\mathsf L_H(x)$ is finite and non-empty for every $x \in H \setminus H^\times$.

With this in mind, assume $H$ is a reduced BF-monoid, so that $\mathsf L_H(x)$ is finite and non-empty for all $x \in H$. Accordingly, pick $\bar x \in H \setminus \{1_H\}$, and let $n_{\bar x}$ be the maximum of $\mathsf L_H(\bar x)$. If $m$ be an integer $\ge 1 + n_x$ and $\bar x = x_1 \cdots x_m$ for some $x_1, \ldots, x_m \in H$, then $$ \mathsf L_H(x_1) + \cdots + \mathsf L_H(x_m) \subseteq \mathsf L_H(\bar x), $$ and hence $$ \max \mathsf L_H(x_1) + \cdots + \max \mathsf L_H(x_m) \le \max \mathsf L_H(\bar x) = n_{\bar x}, $$ which is possible only if $\mathsf L_H(x_i) = \{0\}$, and hence $x_i = 1_H$, for some $i \in [\![1,m]\!]$. []