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Benjamin Steinberg's example is neat and brilliant. Here is a different construction, which, though blatantly complicate, could help with related questions (I had had it in mind for days, but for some reason I couldn't make it work until minutes ago...).

To start with, let $M$ be an additively written monoid (either commutative or not). I'll denote by $\mathcal P(M)$ the extended power monoid of $M$, that is, the monoid obtained by endowing the set of all subsets of $M$ with the binary operation $$ (X, Y) \mapsto X+Y := \{x+y: x \in X \text{ and }y \in Y\}, $$ and by $\mathcal P_{\rm fin}(M)$ the submonoid of $\mathcal P(M)$ consisting of all non-empty finite subsets of $M$ (namely, the power monoid of $M$).

Accordingly, let $\mathbb N = (\mathbf N, +)$ be the additive monoid of non-negative integers and $G$ an additively written abelian group of cardinality $\kappa$, and take $H$ to be the smallest submonoid of $\mathcal P(G \times \mathbb N)$ containing all $1$-element subsets of $G \times \{0\}$, as well as all sets of the form $G \times X$ for which $X$ is a finite subset of $\mathbf N$ with $0 \in X$ and $|X| \ge 2$. Then $H$ is a commutative monoid with $$H^\times = \bigl\{\{(g,0)\bigr\}: g \in G\} \simeq_{\sf Grp} G \quad\text{and}\quad \mathcal A(H) = \{G \times X: 0 \in X \in \mathcal A(\mathcal P_{\rm fin}(\mathbb N))\},$$ and actually a (commutative) BF-monoid, as it follows from considering that $H$ is unit-cancellative and the function $H \to \mathbf N: (A, B) \mapsto |B| - 1$ is a length function (here we use a very basic result from additive number theory, i.e., that $|X+Y| \ge |X| + |Y| - 1$ for all non-empty $X, Y \subseteq \mathbf N$). Moreover, $A + U = A$ for every $U \in H^\times$ and $A \in \mathcal A(H)$, since $\{(g,0)\} + G \times X = G \times X$ for every $X \in \mathcal P_{\rm fin}(\mathbb N)$.

Notes. A monoid $M$ is called unit-cancellative provided that $xy = x$ or $yx = x$ for some $x, y \in M$ only if $y \in M^\times$, and a function $\lambda: M \to \mathbf N$ is a length function if $\lambda(x) < \lambda(y)$ for all $x, y \in M$ for which $y = uxv$ for some $u, v \in M$ with $u \notin M^\times$ or $v \notin M^\times$. If $M$ is unit-cancellative, then it's possible to prove that $M$ is BF iff $M$ has a length function (if requested, I can give a reference).

Benjamin Steinberg's example is neat and brilliant. Here is a different construction, which, though blatantly complicate, could help with related questions (I had had it in mind for days, but for some reason I couldn't make it work until minutes ago...).

To start with, let $M$ be an additively written monoid (either commutative or not). I'll denote by $\mathcal P(M)$ the extended power monoid of $M$, that is, the monoid obtained by endowing the set of all subsets of $M$ with the binary operation $$ (X, Y) \mapsto X+Y := \{x+y: x \in X \text{ and }y \in Y\}, $$ and by $\mathcal P_{\rm fin}(M)$ the submonoid of $\mathcal P(M)$ consisting of all non-empty finite subsets of $M$ (namely, the power monoid of $M$).

Accordingly, let $\mathbb N = (\mathbf N, +)$ be the additive monoid of non-negative integers and $G$ an additively written abelian group of cardinality $\kappa$, and take $H$ to be the smallest submonoid of $\mathcal P(G \times \mathbb N)$ containing all $1$-element subsets of $G \times \{0\}$, as well as all sets of the form $G \times X$ for which $X$ is a finite subset of $\mathbf N$ with $0 \in X$ and $|X| \ge 2$. Then $H$ is a commutative monoid with $$H^\times = \bigl\{\{(g,0)\bigr\}: g \in G\} \simeq_{\sf Grp} G \quad\text{and}\quad \mathcal A(H) = \{G \times X: 0 \in X \in \mathcal A(\mathcal P_{\rm fin}(\mathbb N))\},$$ and actually a (commutative) BF-monoid, as it follows from considering that $H$ is unit-cancellative and the function $H \to \mathbf N: (A, B) \mapsto |B| - 1$ is a length function (here we use a very basic result from additive number theory, i.e., that $|X+Y| \ge |X| + |Y| - 1$ for all non-empty $X, Y \subseteq \mathbf N$). Moreover, $A + U = A$ for every $U \in H^\times$ and $A \in \mathcal A(H)$, since $\{(g,0)\} + G \times X = G \times X$ for every $X \in \mathcal P_{\rm fin}(\mathbb N)$.

Notes. A monoid $M$ is called unit-cancellative provided that $xy = x$ or $yx = x$ for some $x, y \in M$ only if $y \in M^\times$, and a function $\lambda: M \to \mathbf N$ is a length function if $\lambda(x) < \lambda(y)$ for all $x, y \in M$ for which $y = uxv$ for some $u, v \in M$ with $u \notin M^\times$ or $v \notin M^\times$. If $M$ is unit-cancellative, then it's possible to prove that $M$ is BF iff $M$ has a length function (if requested, I can give a reference).

Benjamin Steinberg's example is neat and brilliant. Here is a different construction, which, though blatantly complicate, could help with related questions (I had had it in mind for days, but for some reason I couldn't make it work until minutes ago...).

To start with, let $M$ be an additively written monoid (either commutative or not). I'll denote by $\mathcal P_{\rm fin}(M)$ the power monoid of $M$, that is, the monoid obtained by endowing the set of all non-empty finite subsets of $M$ with the binary operation $$ (X, Y) \mapsto X+Y := \{x+y: x \in X \text{ and }y \in Y\}. $$ Accordingly, let $\mathbb N = (\mathbf N, +)$ be the additive monoid of non-negative integers and $G$ an additively written abelian group of cardinality $\kappa$, and take $H$ to be the smallest submonoid of $\mathcal P_{\rm fin}(G \times \mathbb N)$ containing all $1$-element subsets of $G \times \{0\}$, as well as all sets of the form $G \times X$ for which $X$ is a finite subset of $\mathbf N$ with $0 \in X$ and $|X| \ge 2$. Then $H$ is a commutative monoid with $$H^\times = \bigl\{\{(g,0)\bigr\}: g \in G\} \simeq_{\sf Grp} G \quad\text{and}\quad \mathcal A(H) = \{G \times X: 0 \in X \in \mathcal A(\mathcal P_{\rm fin}(\mathbb N))\},$$ and actually a (commutative) BF-monoid, as it follows from considering that $H$ is unit-cancellative and the function $H \to \mathbf N: (A, B) \mapsto |B| - 1$ is a length function (here we use a very basic result from additive number theory, i.e., that $|X+Y| \ge |X| + |Y| - 1$ for all non-empty $X, Y \subseteq \mathbf N$). Moreover, $A + U = A$ for every $U \in H^\times$ and $A \in \mathcal A(H)$, since $\{(g,0)\} + G \times X = G \times X$ for every $X \in \mathcal P_{\rm fin}(\mathbb N)$.

Notes. A monoid $M$ is called unit-cancellative provided that $xy = x$ or $yx = x$ for some $x, y \in M$ only if $y \in M^\times$, and a function $\lambda: M \to \mathbf N$ is a length function if $\lambda(x) < \lambda(y)$ for all $x, y \in M$ for which $y = uxv$ for some $u, v \in M$ with $u \notin M^\times$ or $v \notin M^\times$. If $M$ is unit-cancellative, then it's possible to prove that $M$ is BF iff $M$ has a length function (if requested, I can give a reference).

Benjamin Steinberg's example is neat and brilliant. Here is a different construction, which, though blatantly complicate, could help with related questions (I had had it in mind for days, but for some reason I couldn't make it work until minutes ago...).

To start with, let $M$ be an additively written monoid (either commutative or not). I'll denote by $\mathcal P(M)$ the extended power monoid of $M$, that is, the monoid obtained by endowing the set of all subsets of $M$ with the binary operation $$ (X, Y) \mapsto X+Y := \{x+y: x \in X \text{ and }y \in Y\}, $$ and by $\mathcal P_{\rm fin}(M)$ the submonoid of $\mathcal P(M)$ consisting of all non-empty finite subsets of $M$ (namely, the power monoid of $M$).

Accordingly, let $\mathbb N = (\mathbf N, +)$ be the additive monoid of non-negative integers and $G$ an additively written abelian group of cardinality $\kappa$, and take $H$ to be the smallest submonoid of $\mathcal P(G \times \mathbb N)$ containing all $1$-element subsets of $G \times \{0\}$, as well as all sets of the form $G \times X$ for which $X$ is a finite subset of $\mathbf N$ with $0 \in X$ and $|X| \ge 2$. Then $H$ is a commutative monoid with $$H^\times = \bigl\{\{(g,0)\bigr\}: g \in G\} \simeq_{\sf Grp} G \quad\text{and}\quad \mathcal A(H) = \{G \times X: 0 \in X \in \mathcal A(\mathcal P_{\rm fin}(\mathbb N))\},$$ and actually a (commutative) BF-monoid, as it follows from considering that $H$ is unit-cancellative and the function $H \to \mathbf N: (A, B) \mapsto |B| - 1$ is a length function (here we use a very basic result from additive number theory, i.e., that $|X+Y| \ge |X| + |Y| - 1$ for all non-empty $X, Y \subseteq \mathbf N$). Moreover, $A + U = A$ for every $U \in H^\times$ and $A \in \mathcal A(H)$, since $\{(g,0)\} + G \times X = G \times X$ for every $X \in \mathcal P_{\rm fin}(\mathbb N)$.

Notes. A monoid $M$ is called unit-cancellative provided that $xy = x$ or $yx = x$ for some $x, y \in M$ only if $y \in M^\times$, and a function $\lambda: M \to \mathbf N$ is a length function if $\lambda(x) < \lambda(y)$ for all $x, y \in M$ for which $y = uxv$ for some $u, v \in M$ with $u \notin M^\times$ or $v \notin M^\times$. If $M$ is unit-cancellative, then it's possible to prove that $M$ is BF iff $M$ has a length function (if requested, I can give a reference).

Benjamin Steinberg's example is neat and brilliant. Here is a different construction, which, though more complicate, could help with related questions.

To start with, let $M$ be an additively written monoid (either commutative or not). I'll denote by $\mathcal P_{\rm fin}(M)$ the power monoid of $M$, that is, the monoid obtained by endowing the set of all non-empty finite subsets of $M$ with the binary operation $$ (X, Y) \mapsto X+Y := \{x+y: x \in X \text{ and }y \in Y\}. $$ Accordingly, let $\mathbb N = (\mathbf N, +)$ be the additive monoid of non-negative integers and $G$ an additively written abelian group of cardinality $\kappa$, and take $H$ to be the smallest submonoid of $\mathcal P_{\rm fin}(G \times \mathbb N)$ containing all $1$-element subsets of $G \times \{0\}$, as well as all sets of the form $G \times X$ for which $X$ is a finite subset of $\mathbf N$ with $0 \in X$ and $|X| \ge 2$. Then $H$ is a commutative monoid with $$H^\times = \bigl\{\{(g,0)\bigr\}: g \in G\} \simeq_{\sf Grp} G \quad\text{and}\quad \mathcal A(H) = \{G \times X: 0 \in X \in \mathcal A(\mathcal P_{\rm fin}(\mathbb N))\},$$ and actually a (commutative) BF-monoid, as it follows from considering that $H$ is unit-cancellative and the function $H \to \mathbf N: (A, B) \mapsto |B| - 1$ is a length function (here we use a very basic result from additive number theory, i.e., that $|X+Y| \ge |X| + |Y| - 1$ for all non-empty $X, Y \subseteq \mathbf N$). Moreover, $A + U = A$ for every $U \in H^\times$ and $A \in \mathcal A(H)$, since $\{(g,0)\} + G \times X = G \times X$ for every $X \in \mathcal P_{\rm fin}(\mathbb N)$.

Notes. A monoid $M$ is called unit-cancellative provided that $xy = x$ or $yx = x$ for some $x, y \in M$ only if $y \in M^\times$, and a function $\lambda: M \to \mathbf N$ is a length function if $\lambda(x) < \lambda(y)$ for all $x, y \in M$ for which $y = uxv$ for some $u, v \in M$ with $u \notin M^\times$ or $v \notin M^\times$. If $M$ is unit-cancellative, then it's possible to prove that $M$ is BF iff $M$ has a length function (if requested, I can give a reference).

Benjamin Steinberg's example is neat and brilliant. Here is a different construction, which, though blatantly complicate, could help with related questions (I had had it in mind for days, but for some reason I couldn't make it work until minutes ago...).

To start with, let $M$ be an additively written monoid (either commutative or not). I'll denote by $\mathcal P_{\rm fin}(M)$ the power monoid of $M$, that is, the monoid obtained by endowing the set of all non-empty finite subsets of $M$ with the binary operation $$ (X, Y) \mapsto X+Y := \{x+y: x \in X \text{ and }y \in Y\}. $$ Accordingly, let $\mathbb N = (\mathbf N, +)$ be the additive monoid of non-negative integers and $G$ an additively written abelian group of cardinality $\kappa$, and take $H$ to be the smallest submonoid of $\mathcal P_{\rm fin}(G \times \mathbb N)$ containing all $1$-element subsets of $G \times \{0\}$, as well as all sets of the form $G \times X$ for which $X$ is a finite subset of $\mathbf N$ with $0 \in X$ and $|X| \ge 2$. Then $H$ is a commutative monoid with $$H^\times = \bigl\{\{(g,0)\bigr\}: g \in G\} \simeq_{\sf Grp} G \quad\text{and}\quad \mathcal A(H) = \{G \times X: 0 \in X \in \mathcal A(\mathcal P_{\rm fin}(\mathbb N))\},$$ and actually a (commutative) BF-monoid, as it follows from considering that $H$ is unit-cancellative and the function $H \to \mathbf N: (A, B) \mapsto |B| - 1$ is a length function (here we use a very basic result from additive number theory, i.e., that $|X+Y| \ge |X| + |Y| - 1$ for all non-empty $X, Y \subseteq \mathbf N$). Moreover, $A + U = A$ for every $U \in H^\times$ and $A \in \mathcal A(H)$, since $\{(g,0)\} + G \times X = G \times X$ for every $X \in \mathcal P_{\rm fin}(\mathbb N)$.

Notes. A monoid $M$ is called unit-cancellative provided that $xy = x$ or $yx = x$ for some $x, y \in M$ only if $y \in M^\times$, and a function $\lambda: M \to \mathbf N$ is a length function if $\lambda(x) < \lambda(y)$ for all $x, y \in M$ for which $y = uxv$ for some $u, v \in M$ with $u \notin M^\times$ or $v \notin M^\times$. If $M$ is unit-cancellative, then it's possible to prove that $M$ is BF iff $M$ has a length function (if requested, I can give a reference).