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The answer is negative for the class of countable Boolean algebras. The reference is Jussi Ketonen's "The Structure of Countable Boolean Algebras" (Annals of Mathematics [Second Series], Vol. 108, 1978, No. 1, pp. 41-89). There, Ketonen shows any countable commutative semigroup can be embedded into the monoid of countable Boolean algebras. The proof of this is rather involved.

The answer is positive for the class of linear orders (replacing product with concatenation). Lindenbaum showed for any linear orders $y$ and $z$, if $y$ is an initial segment of $z$ and $z$ is an end segment of $y$, then $y \cong z$. Taking $x+x$ for $y$ and $x = x+x+x$ for $z$ suffices. A reference is Joseph Rosenstein's "Linear Orderings" (Academic Press Inc., New York, 1982, p.22). The proof of this is rather straightforward.

The answer is negative for the class of countable Boolean algebras. The reference is Jussi Ketonen's "The Structure of Countable Boolean Algebras" (Annals of Mathematics [Second Series], Vol. 108, 1978, No. 1, pp. 41-89). There, Ketonen shows any countable commutative semigroup can be embedded into the monoid of countable Boolean algebras. The proof of this is rather involved.

The answer is positive for the class of linear orders (replacing product with concatenation). Lindenbaum showed for any linear orders $y$ and $z$, if $y$ is an initial segment of $z$ and $z$ is an end segment of $y$, then $y \cong z$. Taking $x+x$ for $y$ and $x = x+x+x$ for $z$ suffices. A reference is Joseph Rosenstein's "Linear Orderings" (Academic Press Inc., New York, 1982, p.22). The proof of this is rather straightforward.

The answer is negative for the class of countable Boolean algebras. The reference is Jussi Ketonen's "The Structure of Countable Boolean Algebras" (Annals of Mathematics [Second Series], Vol. 108, 1978, No. 1, pp. 41-89). There, Ketonen shows any countable commutative semigroup can be embedded into the monoid of countable Boolean algebras. The proof of this is rather involved.

The answer is positive for the class of linear orders (replacing product with concatenation). Lindenbaum showed for any linear orders $y$ and $z$, if $y$ is an initial segment of $z$ and $z$ is an end segment of $y$, then $y \cong z$. Taking $x+x$ for $y$ and $x = x+x+x$ for $z$ suffices. A reference is Joseph Rosenstein's "Linear Orderings" (Academic Press Inc., New York, 1982, p.22). The proof of this is rather straightforward.

The answer is negative for the class of countable Boolean algebras. The reference is Jussi Ketonen's "The Structure of Countable Boolean Algebras" (Annals of Mathematics [Second Series], Vol. 108, 1978, No. 1, pp. 41-89). There, Ketonen shows any countable commutative semigroup can be embedded into the monoid of countable Boolean algebras. The proof of this is rather involved.

The answer is positive for the class of linear orders (replacing product with concatenation). Lindenbaum showed for any linear orders $y$ and $z$, if $y$ is an initial segment of $z$ and $z$ is an end segment of $y$, then $y \cong z$. Taking $x+x$ for $y$ and $x = x+x+x$ for $z$ suffices. A reference is Joseph Rosenstein's "Linear Orderings" (Academic Press Inc., New York, 1982, p.22). The proof of this is rather straightforward.

The answer is negative for the class of countable Boolean algebras. The reference is Jussi Ketonen's "The Structure of Countable Boolean Algebras" (Annals of Mathematics [Second Series], Vol. 108, 1978, No. 1, pp. 41-89). There, Ketonen shows any countable commutative semigroup can be embedded into the monoid of countable Boolean algebras. The proof of this is rather involved.

The answer is positive for the class of linear orders (replacing product with concatenation). Lindenbaum showed for any linear orders $y$ and $z$, if $y$ is an initial segment of $z$ and $z$ is an end segment of $y$, then $x \cong y$. Taking $x+x$ for $y$ and $x = x+x+x$ for $z$ suffices. A reference is Joseph Rosenstein's "Linear Orderings" (Academic Press Inc., New York, 1982, p.22). The proof of this is rather straightforward.

The answer is negative for the class of countable Boolean algebras. The reference is Jussi Ketonen's "The Structure of Countable Boolean Algebras" (Annals of Mathematics [Second Series], Vol. 108, 1978, No. 1, pp. 41-89). There, Ketonen shows any countable commutative semigroup can be embedded into the monoid of countable Boolean algebras. The proof of this is rather involved.

The answer is positive for the class of linear orders (replacing product with concatenation). Lindenbaum showed for any linear orders $y$ and $z$, if $y$ is an initial segment of $z$ and $z$ is an end segment of $y$, then $y \cong z$. Taking $x+x$ for $y$ and $x = x+x+x$ for $z$ suffices. A reference is Joseph Rosenstein's "Linear Orderings" (Academic Press Inc., New York, 1982, p.22). The proof of this is rather straightforward.