Let $R = \Bbb{Z}^3$ be the usual direct product ring, and let $S = \{ (k^a, k^b, k^c) : k \in \Bbb{Z} - \{0\}\}$ be a subsemigroup of $R$. Then do the cosets $aS$ partition $R$? Let $R^* = S^{-1}R$ be $R$. Then $R^*$ is a ring containing an isomorphic copy of $R$, with multiplicative group of units equal to the image of $S$ unioned with the inverse elements $S^{-1}$. Call the multiplicative group $U$. Then since $1 \in U, \ \bigcup_{a \in R^*} a U = R^*$ and also if $a U \cap b U \neq \varnothing$ then $as = bs' \implies a = b s' s^{-1}$, so $a U = b U$. Thus the cosets $aU$ partition $R^*$.
Does this imply that cosets of the semigroup $S$ partition $R$?
I would say yes, since if $aS \cap bS \neq \varnothing$ then $aU = bU \implies ?$
where I'm stuck.
Let $b = (k_0^a, k_0^b, k_0^c)$ for some $k_0 \in \Bbb{Z} - \{0\}$ not equal to a unit ($(\pm 1, \dots)$). Then $ab S \subsetneq a S$. Let $x = (k_1^a, \dots, k_1^c) \in S$ be such that $k_1 \lt k_0$ i.e. choose them both in that way. Then for any non-zero $a$ not a zero divisor, $ \ ax = ab s \implies a(x -bs) = 0 \implies x = bs$ but that's impossible.
Thus $abS \subsetneq aS$ so their intersection violates the partition def, so cosets of $S$ do not partition $R$.