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Let $S$ be the positive cone of a non-abelian, totally ordered group $(G, \leq)$ all of whose proper quotients are abelian (the existence of which is proved by YCor as part of his construction). In particular, this means that $\leq$ is a total order on $G$ such that $a \leq b$ implies $xay \leq xby$ for all $x, y \in G$, and $S$ is the subsemigroup $\{a \in G: 1_G \lneq a\}$ of $G$ (everything here is written multiplicatively).

Proof. Let $f \colon H \to K$ be a surjective semigroup homomorphism and assume $H$ is duo. If $u, v \in K$, then $u = f(x)$ and $v = f(y)$ for some $x, y \in H$; and since $H$ is duo, there are $a, b \in H$ such that $xy = ax = yb$. It follows that $uv = f(xy) = f(ax) = f(a) u$ and, in a similar way, $uv = v f(b)$. Thus, $K$ is duo. []

For the next results, let me recall that a subset $I$ of a semigroup $H$ is a left (resp., right) ideal if $HI \subseteq I$ (resp., $IH\subseteq I$); some people would also require that $I$ is non-empty, but we don't. In addition, $I$ is a (two-sided) ideal if it is both a left and a right ideal.

Proof. Let $M$ be a minimal ideal of $H$. In a duo semigroup, every left ideal is a right ideal, and vice versa; see, for instance, Excercise 22.4A in Lam's A first course on noncommutative rings (the exercise is about duo rings, but the proof carries over verbatim to duo semigroups). Thus, $M$ is also a (non-empty) minimal right ideal. It follows that $aM = M$ for all $a \in M$ (since $aM$ is itself a non-empty right ideal, and it is contained in $M$), and so $a=ae$ for some $e\in M$. Then $ae^2=ae$ implies (by cancellativity) that $e^2=e$. But an idempotent in a cancellative semigroup must be an identity (exercise). Thus, $H=eH=M$ is a monoid. This yields (by minimality) that, for any $b \in H$, $H = bH = Hb$, that is, the equations $bx = e$ and $yb = e$ are both solvable in $H$. Therefore, $H$ is a group. []

Lemma 4. Every non-empty, subdirectly irreducible semigroup has a (non-empty) minimal ideal.

Proof. Let $H$ be a non-empty, subdirectly irreducible semigroup, and let $\mathfrak I(H)$ be the family of all non-empty ideals of $H$. Clearly, $M := \bigcap \mathfrak I(H)$ is itself an ideal of $H$; and by construction, it is contained in any non-empty ideal. We are left to show that $M \ne \emptyset$.

To this end, denote by $\theta_I$ the Rees congruence on $H$ induced by an ideal $I$ of $H$ (see, for instance, here), and note that $I$ is empty if and only if $\theta_I$ is trivial, that is, $\theta_I = \Delta_H := \{(x,x): x \in H\}$. Since a semigroup is subdirectly irreducible if and only if the intersection of all its non-trivial congruences is non-trivial (see, for instance, p. 765 of the 1944 BAMS paper by Birkhoff cited in the OP), it follows that $\theta_M \ne \Delta_H$ and hence $M \ne \emptyset$. []

Let $S$ be the positive cone of a non-abelian, totally ordered group $(G, \leq)$ all of whose proper quotients are abelian (the existence of such an object is proved by YCor as part of his construction). In particular, this means that $\leq$ is a total order on $G$ with the further property that $a \leq b$ implies $xay \leq xby$ for all $x, y \in G$, and $S$ is the subsemigroup $\{a \in G: 1_G \lneq a\}$ of $G$ (everything here is written multiplicatively).

Proof. Let $f \colon H \to K$ be a surjective semigroup homomorphism, and assume $H$ is duo. If $u, v \in K$, then $u = f(x)$ and $v = f(y)$ for some $x, y \in H$; and since $H$ is duo, there are $a, b \in H$ such that $xy = ax = yb$. It follows that $uv = f(xy) = f(ax) = f(a) u$ and, in a similar way, $uv = v f(b)$. Thus, $K$ is duo. []

For the next results, recall that a subset $I$ of a semigroup $H$ is a left (resp., right) ideal if $HI \subseteq I$ (resp., $IH\subseteq I$); some people would also require that $I$ is non-empty, but we don't. In addition, $I$ is a (two-sided) ideal if it is both a left and a right ideal.

Proof. Let $M$ be a non-empty minimal ideal of $H$. In a duo semigroup, every left ideal is a right ideal, and vice versa; see, for instance, Exercise 22.4A in Lam's A First Course in Noncommutative Rings (the exercise is about duo rings, but the proof carries over verbatim to duo semigroups). Thus, $M$ is also a non-empty minimal right ideal. It follows that $aM = M$ for all $a \in M$ (since $aM$ is itself a non-empty right ideal, and it is contained in $M$), and so $a=ae$ for some $e\in M$. Then $ae^2=ae$ implies (by cancellativity) that $e^2=e$. But an idempotent in a cancellative semigroup must be an identity (exercise). Thus, $H=eH=M$ is a monoid and $e$ is its identity. This yields (by minimality) that, for any $b \in H$, $H = bH = Hb$, that is, the equations $bx = e$ and $yb = e$ are both solvable in $H$. Therefore, $H$ is a group. []

Lemma 4. Every non-empty, subdirectly irreducible semigroup $H$ has a non-empty minimal ideal.

Proof. Let $\mathfrak I(H)$ be the family of all non-empty ideals of $H$. Clearly, $M := \bigcap \mathfrak I(H)$ is itself an ideal of $H$; and by construction, it is contained in any non-empty ideal. We are left to show that $M \ne \emptyset$.

To this end, denote by $\theta_I$ the Rees congruence on $H$ induced by an ideal $I$ of the semigroup itself (see, for instance, here), and note that $I$ is empty if and only if $\theta_I$ is trivial, that is, $\theta_I = \Delta_H := \{(x,x): x \in H\}$. Since a semigroup is subdirectly irreducible if and only if the intersection of all its non-trivial congruences is non-trivial (see, for instance, p. 765 of the 1944 BAMS paper by Birkhoff cited in the OP), it follows that $\theta_M \ne \Delta_H$ and hence $M \ne \emptyset$. []

Proof. Let $M$ be a minimal ideal of $H$. In a duo semigroup, every left ideal is a right ideal, and vice versa; see, for instance, Excercise 22.4A in Lam's A first course on noncommutative rings (the exercise is about duo rings, but the proof carries over verbatim to duo semigroups). Thus, $M$ is also a (non-empty) minimal right ideal. It follows that $aM = M$ for all $a \in M$ (since $aM$ is itself a non-empty right ideal, and it is contained in $M$), and so $a=ae$ for some $e\in M$. Then $ae^2=ae$ implies (by cancellativity) that $e^2=e$. But an idempotent in a cancellative semigroup must be an identity (exercise). Thus, $H=eH=M$ is a monoid. This yields (by minimality) that, for any $b \in H$, $H = bH = Hb$, that is, the equations $bx = e$ and $yb = e$ are both solvable in $H$. Therefore, $H$ is a group.

Proof. Let $M$ be a minimal ideal of $H$. In a duo semigroup, every left ideal is a right ideal, and vice versa; see, for instance, Excercise 22.4A in Lam's A first course on noncommutative rings (the exercise is about duo rings, but the proof carries over verbatim to duo semigroups). Thus, $M$ is also a (non-empty) minimal right ideal. It follows that $aM = M$ for all $a \in M$ (since $aM$ is itself a non-empty right ideal, and it is contained in $M$), and so $a=ae$ for some $e\in M$. Then $ae^2=ae$ implies (by cancellativity) that $e^2=e$. But an idempotent in a cancellative semigroup must be an identity (exercise). Thus, $H=eH=M$ is a monoid. This yields (by minimality) that, for any $b \in H$, $H = bH = Hb$, that is, the equations $bx = e$ and $yb = e$ are both solvable in $H$. Therefore, $H$ is a group. []

Sorry for answering my own question, but YCor's construction in a related thread (here) gave me a lightbulb moment. I can only hope it's not a broken lightbulb.

Let $S$ be the positive cone of a non-abelian linearly orderable group $G$ such that all proper quotients are abelian (the existence of which is proved by YCor as part of his construction); in plain English, this means that there is a total order $\leq$ on $G$ such that $a \leq b$ implies $xay \leq xby$ for all $x, y \in G$, and $S$ is the subsemigroup $\{a \in G: 1_G \lneq a\}$ of $G$ (everything here is written multiplicatively).

Theorem. If $S$ is the positive cone in a non-abelian linearly orderable group with no proper non-abelian quotients, such as the positive cone of the Thompson group $F$, ordered as in YCor's answer here, then at least one factor in any subdirectly irreducible subdirect representation of $S$ is not cancellative.

The proof is not complicated, but is going to demand some work. I will organize it into a series of lemmas.

Lemma 1. $S$ is a duo semigroup, that is, $aS = Sa$ for all $a \in S$.

Lemma 3. If a cancellative duo semigroup $H$ has a (non-empty) minimal ideal, then it is a group.

Proof. Let $M$ be a minimal ideal of $H$. In a duo semigroup, every left ideal is a right ideal, and vice versa; see, for instance, Excercise 22.4A in Lam's A first course (the exercise is about duo rings, but the proof carries over verbatim to duo semigroups). Thus, $M$ is also a non-empty minimal right ideal. So if $a\in M$, then $aM=M$ by minimality, and so $a=ae$ for some $e\in M$. Then $ae^2=ae$ implies that $e^2=e$. But an idempotent in a cancellative semigroup must be an identity (exercise). Thus $H=eH=M$ is a monoid. But then by minimality, we have $cH=H=Hc$ for all $c\in H$, that is the equations $cx=e$ and $yc=e$ have solutions for any $c\in H$. Therefore $H$ is a group.

Proof of Theorem. Suppose for the sake of contradiction that the semigroup $S$ defined at the beginning of this post has a subdirect representation $p \colon S \to \prod_{j \in J} S_j$, where the $S_j$'s are subdirectly irreducible cancellative semigroups. Each of the $S_j$'s is then a homomorphic image of $S$; so, by Lemmas 1 and 2, $S_j$ is a (cancellative) duo semigroup. It follows, by Lemmas 3 and 4, that the $S_j$'s are all groups. This is however impossible, for YCor showed here that $S$ is not a subdirect product of groups, regardless of whether they are directly irreducible or not. []

Sorry for answering my own question, but YCor's construction in a related thread (here) gave me a lightbulb moment. Hopefully, it's not a broken lightbulb.

Let $S$ be the positive cone of a non-abelian, totally ordered group $(G, \leq)$ all of whose proper quotients are abelian (the existence of which is proved by YCor as part of his construction). In particular, this means that $\leq$ is a total order on $G$ such that $a \leq b$ implies $xay \leq xby$ for all $x, y \in G$, and $S$ is the subsemigroup $\{a \in G: 1_G \lneq a\}$ of $G$ (everything here is written multiplicatively).

Theorem. The semigroup $S$ is not a subdirect product of subdirectly irreducible, cancellative semigroups.

The proof is going to demand some work. I will organize it into a series of lemmas.

Lemma 1. The positive cone $H$ of any totally ordered group is a (cancellative) duo semigroup, that is, $aH = Ha$ for all $a \in H$.

For the next results, let me recall that a subset $I$ of a semigroup $H$ is a left (resp., right) ideal if $HI \subseteq I$ (resp., $IH\subseteq I$); some people would also require that $I$ is non-empty, but we don't. In addition, $I$ is a (two-sided) ideal if it is both a left and a right ideal.

Lemma 3. If a cancellative duo semigroup $H$ has a non-empty minimal ideal, then it is a group.

Proof. Let $M$ be a minimal ideal of $H$. In a duo semigroup, every left ideal is a right ideal, and vice versa; see, for instance, Excercise 22.4A in Lam's A first course on noncommutative rings (the exercise is about duo rings, but the proof carries over verbatim to duo semigroups). Thus, $M$ is also a (non-empty) minimal right ideal. It follows that $aM = M$ for all $a \in M$ (since $aM$ is itself a non-empty right ideal, and it is contained in $M$), and so $a=ae$ for some $e\in M$. Then $ae^2=ae$ implies (by cancellativity) that $e^2=e$. But an idempotent in a cancellative semigroup must be an identity (exercise). Thus, $H=eH=M$ is a monoid. This yields (by minimality) that, for any $b \in H$, $H = bH = Hb$, that is, the equations $bx = e$ and $yb = e$ are both solvable in $H$. Therefore, $H$ is a group.

Proof of Theorem. Suppose for the sake of contradiction that the semigroup $S$ defined at the beginning of this post has a subdirect representation $p \colon S \to \prod_{j \in J} S_j$, where all the $S_j$'s are subdirectly irreducible cancellative semigroups. Each of the $S_j$'s is then a homomorphic image of $S$; so, by Lemmas 1 and 2, $S_j$ is a (cancellative) duo semigroup. It follows, by Lemmas 3 and 4, that the $S_j$'s are groups. This is however impossible, for YCor showed here that $S$ is not a subdirect product of groups. []