**I am hoping that there is a generalization of Birkhoff's theorem that applies not only to algebras (in the sense of Birkhoff's paper) but also to algebras that satisfy certain equational identities (such as cancellative semigroups).**<p>

There is a version of Birkhoff's Theorem that applies to quasivarieties, such as the quasivariety of cancellative semigroups, but it doesn't give you what you are asking for here. Namely, if $\mathcal{Q}$ is a quasivariety and $\mathbf{A}\in \mathcal{Q}$, then call $\theta$ a $\mathcal{Q}$-congruence on $\mathbf{A}$ if $\mathbf{A}/\theta\in \mathcal{Q}$. The $\mathcal{Q}$-congruences on $\mathbf{A}$ form an algebraic lattice, and that is enough for the
'$\mathcal{Q}$-version' of Birkhoff's Theorem to hold: every algebra in $\mathcal{Q}$ is a subdirect product of '$\mathcal{Q}$-subdirectly irreducibles'. An algebra $\mathbf{S}\in \mathcal{Q}$ is $\mathcal{Q}$-subdirectly irreducible if it has a least nonzero $\mathcal{Q}$-congruence. In particular, every cancellative semigroup is a subdirect product of cancellative semigroups that are subdirectly irreducible relative to the class of cancellative semigroups.<p>

But a $\mathcal{Q}$-subdirectly irreducible might not be subdirectly irreducible in the absolute sense, so this doesn't answer the question that you asked. 
[For example, if $\mathcal{Q}$ is the quasivariety of torsion-free abelian groups, then every member of $\mathcal{Q}$
is a subdirect product of $\mathcal{Q}$-subdirectly irreducibles. But no torsion-free abelian group is subdirectly irreducible in the absolute sense, so the members of $\mathcal{Q}$ are not subdirect products of (absolutely) subdirectly irreducible torsion-free abelian groups.]


Let me at least give an affirmative answer to the commutative version of your question.<p>

**Claim.** If $S$ is a commutative cancellative group, 
then $S$ is a subdirect product of subdirectly irreducible cancellative semigroups. (In fact, $S$ is a subdirect product of subdirectly irreducible abelian <b>groups</b> considered as semigroups.)<p>

**Reasoning.** Embed $S$ in its universal group $U$, which is an abelian group.
For each $a\neq b$ in $S$ ($\subseteq U$), choose a group congruence $\theta$ on $U$ that is maximal for $(a,b)\notin \theta$. The group $U/\theta$ is a subdirectly irreducible abelian group, hence it is isomorphic to a subgroup of some Pr&uuml;fer group $\mathbb Z_{p^{\infty}}$. The composite map
$f_{a,b}\colon S\to U\to U/\theta\to \mathbb Z_{p^{\infty}}$
separates $a$ and $b$ and has image $f_{a,b}(S)$
that is a subsemigroup of
$\mathbb Z_{p^{\infty}}$. Every subsemigroup of
$\mathbb Z_{p^{\infty}}$ is a subdirectly irreducible group
that is also subdirectly irreducible as a semigroup.
Thus, $\prod f_{a,b}\colon S\to \prod f_{a,b}(S)$ represents
$S$ as a subdirect product of subdirectly irreducible abelian groups. \\\\\\<p>

---------------

**Edit.**
Let me respond to a request in the comments by giving a reference for the relative version of Birkhoff's Theorem.
See Theorem 1.1 of <p>

[Finite basis theorems for relatively congruence-distributive quasivarieties][1]<br>
    Don Pigozzi<br> 
    Trans. Amer. Math. Soc. 310 (1988), 499-533. 


  [1]: https://www.ams.org/journals/tran/1988-310-02/S0002-9947-1988-0946222-1?active=current