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Figured it out (sorry for answering my own question). I'll prove the following:

Lemma. Let $H$ be a linearly orderable monoid and $R$ a domain whose group of units is trivial (e.g., $R$ could be the ring of integers mod $2$). Then $H$ embeds as a divisor-closed submonoid into the multiplicative monoid of the monoid ring $R[H]$.

This will show that the answer to the question in the OP is "For all $G$", since $\mathscr B(G)$ is a torsion-free, cancellative, commutative monoid, and hence is linearly orderable.

Proof of the lemma. Let $\preceq$ be a total order on $H$ for which $(H, \preceq)$ is a linearly ordered monoid, that is, $xz \prec yz$ and $zx \prec zy$ for all $x, y, z \in H$ with $x \prec y$ (as usual, $u \prec v$ means $u \preceq v$ and $u \ne v$). Moreover, let $\delta$ be the canonical embedding $H \to R[H]$, so that $\delta(x)$ is, for every $x \in H$, a Kronecker delta $H \to R$ centered at $x$, which I'll rather denote by $\delta_x$.

Now pick $x \in H$, and assume $\delta_x = f\ast g$ for some $f, g \in R[H]$, with $\ast$ being the multiplication in $R[H]$. I claim that $|{\rm supp}(f)| = |{\rm supp}(g)| = 1$. Indeed, it is clear that $f$ and $g$ cannot be identically zero, so the support of each of them is non-empty. Accordingly, let $y_f$ and $z_f$ be, respectively, the minimum and the maximum of $S_f := {\rm supp}(f)$ relative to the order $\preceq$ (which exist because $\preceq$ is total and $S_f$ is not only non-empty, but also finite); define $y_g$ and $z_g$ in a similar way (only with $g$ in lieu of $f$). Then set $y := y_f y_g$ and $z := z_f z_g$. We have $$ \delta_x(y) = \sum_{uv=y} f(u) g(v) = f(y_f) g(y_g) \ne 0_R.$$ To see this, note that, if $u \prec y_f$ or $v \preceq y_g$, then $f(u) g(v) = 0_R$, and on the other hand, if $y_f \preceq u$ and $y_g \preceq v$, then $y = y_f y_g \preceq uv$, with equality iff $u = y_f$ and $v = y_g$ (by the assumption that $H$ is linearly ordered by $\preceq$). Then the rest is trivial, because $y_f \in {\rm supp}(f)$ and $y_g \in {\rm supp}(g)$ give $f(y_f) g(y_g) \ne 0_R$ (by the fact that $R$ is a domain).

To wit, we have shown that $y$ is in the support of $\delta_x$, and so is $z$ by an analogous argument. It follows $x = y = z$, which is, however, possible only if $y_f = z_f$ and $y_g = z_g$ (by construction, $y_f \preceq z_f$ and $y_g \preceq z_g$, and we can't have $y_f \neq z_f$ or $y_g \ne z_g$, otherwise $H$ being linearly ordered by $\preceq$ would yield a contradiction).

In other words, $f = a\delta_u$ and $g = b \delta_v$ for some $a, b \in R$ and $u, v \in H$ such that $ab= 1_R$. But $R^\times$ is trivial (by hypothesis), and therefore $a = b = 1_R$. So $f, g \in \delta(H)$, and we are done. []

Figured it out (sorry for answering my own question). I'll prove the following:

Lemma. Let $H$ be a linearly orderable monoid and $R$ a domain whose group of units is trivial. Then $H$ embeds as a divisor-closed submonoid into the multiplicative monoid of the monoid ring $R[H]$.

This will show that the answer to the question in the OP is "For all $G$", since $\mathscr B(G)$ is a torsion-free, cancellative, commutative monoid, and hence is linearly orderable.

Proof of the lemma. Let $\preceq$ be a total order on $H$ for which $(H, \preceq)$ is a linearly ordered monoid, that is, $xz \prec yz$ and $zx \prec zy$ for all $x, y, z \in H$ with $x \prec y$ (as usual, $u \prec v$ means $u \preceq v$ and $u \ne v$). Moreover, let $\delta$ be the canonical embedding $H \to R[H]$, so that $\delta(x)$ is, for every $x \in H$, a Kronecker delta $H \to R$ centered at $x$, which I'll rather denote by $\delta_x$.

Now pick $x \in H$, and assume $\delta_x = f\ast g$ for some $f, g \in R[H]$, with $\ast$ being the multiplication in $R[H]$. I claim that $|{\rm supp}(f)| = |{\rm supp}(g)| = 1$. Indeed, it is clear that $f$ and $g$ cannot be identically zero, so the support of each of them is non-empty. Accordingly, let $y_f$ and $z_f$ be, respectively, the minimum and the maximum of $S_f := {\rm supp}(f)$ relative to the order $\preceq$ (which exist because $\preceq$ is total and $S_f$ is not only non-empty, but also finite); define $y_g$ and $z_g$ in a similar way (only with $g$ in lieu of $f$). Then set $y := y_f y_g$ and $z := z_f z_g$. We have $$ \delta_x(y) = \sum_{uv=y} f(u) g(v) = f(y_f) g(y_g) \ne 0_R.$$ To see this, note that, if $u \prec y_f$ or $v \prec y_g$, then $f(u) g(v) = 0_R$, and on the other hand, if $y_f \preceq u$ and $y_g \preceq v$, then $y = y_f y_g \preceq uv$, with equality iff $u = y_f$ and $v = y_g$ (by the assumption that $H$ is linearly ordered by $\preceq$). Then the rest is trivial, because $y_f \in {\rm supp}(f)$ and $y_g \in {\rm supp}(g)$ give $f(y_f) g(y_g) \ne 0_R$ (by the fact that $R$ is a domain).

To wit, we have shown that $y$ is in the support of $\delta_x$, and so is $z$ by an analogous argument. It follows $x = y = z$, which is, however, possible only if $y_f = z_f$ and $y_g = z_g$ (by construction, $y_f \preceq z_f$ and $y_g \preceq z_g$, and we can't have $y_f \neq z_f$ or $y_g \ne z_g$, otherwise $H$ being linearly ordered by $\preceq$ would yield a contradiction).

In other words, $f = a\delta_u$ and $g = b \delta_v$ for some $a, b \in R$ and $u, v \in H$ such that $ab= 1_R$. But $R^\times$ is trivial (by hypothesis), and therefore $a = b = 1_R$. So $f, g \in \delta(H)$, and we are done. []

In the last line of the proof, we use that domains are, of course, Dedekind-finite rings, so that a product is a unit iff so are all the factors (see here).

Figured it out (sorry for answering my own question). I'll prove the following:

Lemma. Let $H$ be a linearly orderable monoid and $R$ a domain whose group of units is trivial (e.g., $R$ could be the ring of integers mod $2$). Then $H$ embeds as a divisor-closed submonoid into the multiplicative monoid of the monoid ring $R[H]$.

This will provide an affirmative answer to the question in the OP, since $\mathscr B(G)$ is a torsion-free, cancellative, commutative monoid, and hence is linearly orderable.

Proof of the lemma. Let $\preceq$ be a total order on $H$ for which $(H, \preceq)$ is a linearly ordered monoid, that is, $xz \prec yz$ and $zx \prec zy$ for all $x, y, z \in H$ with $x \prec y$ (as usual, $u \prec v$ means $u \preceq v$ and $u \ne v$). Moreover, let $\delta$ be the canonical embedding $H \to R[H]$, so that $\delta(x)$ is, for every $x \in H$, a Kronecker delta $H \to R$ centered at $x$, which I'll rather denote by $\delta_x$.

Now pick $x \in H$, and assume $\delta_x = f\ast g$ for some $f, g \in R[H]$, with $\ast$ being the multiplication in $R[H]$. I claim that $|{\rm supp}(f)| = |{\rm supp}(g)| = 1$. Indeed, it is clear that $f$ and $g$ cannot be identically zero, so the support of each of them is non-empty. Accordingly, let $y_f$ and $z_f$ be, respectively, the minimum and the maximum of $S_f := {\rm supp}(f)$ relative to the order $\preceq$ (which exist because $\preceq$ is total and $S_f$ is not only non-empty, but also finite); define $y_g$ and $z_g$ in a similar way (only with $g$ in lieu of $f$). Then set $y := y_f y_g$ and $z := z_f z_g$. We have $$ \delta_x(y) = \sum_{uv=y} f(u) g(v) = f(y_f) g(y_g) \ne 0_R.$$ To see this, note that, if $u \prec y_f$ or $v \preceq y_g$, then $f(u) g(v) = 0_R$, and on the other hand, if $y_f \preceq u$ and $y_g \preceq v$, then $y = y_f y_g \preceq uv$, with equality iff $u = y_f$ and $v = y_g$ (by the assumption that $H$ is linearly ordered by $\preceq$). Then the rest is trivial, because $y_f \in {\rm supp}(f)$ and $y_g \in {\rm supp}(g)$ give $f(y_f) g(y_g) \ne 0_R$ (by the fact that $R$ is a domain).

To wit, we have shown that $y$ is in the support of $\delta_x$, and so is $z$ by an analogous argument. It follows $x = y = z$, which is, however, possible only if $y_f = z_f$ and $y_g = z_g$ (by construction, $y_f \preceq z_f$ and $y_g \preceq z_g$, and we can't have $y_f \neq z_f$ or $y_g \ne z_g$, otherwise $H$ being linearly ordered by $\preceq$ would yield a contradiction).

In other words, $f = a\delta_u$ and $g = b \delta_v$ for some $a, b \in R$ and $u, v \in H$ such that $ab= 1_R$. But $R^\times$ is trivial (by hypothesis), and therefore $a = b = 1_R$. So $f, g \in \delta(H)$, and we are done. []

Figured it out (sorry for answering my own question). I'll prove the following:

Lemma. Let $H$ be a linearly orderable monoid and $R$ a domain whose group of units is trivial (e.g., $R$ could be the ring of integers mod $2$). Then $H$ embeds as a divisor-closed submonoid into the multiplicative monoid of the monoid ring $R[H]$.

This will show that the answer to the question in the OP is "For all $G$", since $\mathscr B(G)$ is a torsion-free, cancellative, commutative monoid, and hence is linearly orderable.

Proof of the lemma. Let $\preceq$ be a total order on $H$ for which $(H, \preceq)$ is a linearly ordered monoid, that is, $xz \prec yz$ and $zx \prec zy$ for all $x, y, z \in H$ with $x \prec y$ (as usual, $u \prec v$ means $u \preceq v$ and $u \ne v$). Moreover, let $\delta$ be the canonical embedding $H \to R[H]$, so that $\delta(x)$ is, for every $x \in H$, a Kronecker delta $H \to R$ centered at $x$, which I'll rather denote by $\delta_x$.

Now pick $x \in H$, and assume $\delta_x = f\ast g$ for some $f, g \in R[H]$, with $\ast$ being the multiplication in $R[H]$. I claim that $|{\rm supp}(f)| = |{\rm supp}(g)| = 1$. Indeed, it is clear that $f$ and $g$ cannot be identically zero, so the support of each of them is non-empty. Accordingly, let $y_f$ and $z_f$ be, respectively, the minimum and the maximum of $S_f := {\rm supp}(f)$ relative to the order $\preceq$ (which exist because $\preceq$ is total and $S_f$ is not only non-empty, but also finite); define $y_g$ and $z_g$ in a similar way (only with $g$ in lieu of $f$). Then set $y := y_f y_g$ and $z := z_f z_g$. We have $$ \delta_x(y) = \sum_{uv=y} f(u) g(v) = f(y_f) g(y_g) \ne 0_R.$$ To see this, note that, if $u \prec y_f$ or $v \preceq y_g$, then $f(u) g(v) = 0_R$, and on the other hand, if $y_f \preceq u$ and $y_g \preceq v$, then $y = y_f y_g \preceq uv$, with equality iff $u = y_f$ and $v = y_g$ (by the assumption that $H$ is linearly ordered by $\preceq$). Then the rest is trivial, because $y_f \in {\rm supp}(f)$ and $y_g \in {\rm supp}(g)$ give $f(y_f) g(y_g) \ne 0_R$ (by the fact that $R$ is a domain).

To wit, we have shown that $y$ is in the support of $\delta_x$, and so is $z$ by an analogous argument. It follows $x = y = z$, which is, however, possible only if $y_f = z_f$ and $y_g = z_g$ (by construction, $y_f \preceq z_f$ and $y_g \preceq z_g$, and we can't have $y_f \neq z_f$ or $y_g \ne z_g$, otherwise $H$ being linearly ordered by $\preceq$ would yield a contradiction).

In other words, $f = a\delta_u$ and $g = b \delta_v$ for some $a, b \in R$ and $u, v \in H$ such that $ab= 1_R$. But $R^\times$ is trivial (by hypothesis), and therefore $a = b = 1_R$. So $f, g \in \delta(H)$, and we are done. []

Figured it out (sorry for answering my own question). I'll prove the following:

Lemma. Let $H$ be a linearly orderable monoid and $R$ a domain whose group of units is trivial (e.g., $R$ could be the ring of integers mod $2$). Then $H$ embeds as a divisor-closed submonoid into the multiplicative monoid of the monoid ring $R[H]$.

This will provide an affirmative answer to the question in the OP, since $\mathscr B(G)$ is a torsion-free commutative monoid, and hence is linearly orderable.

Proof of the lemma. Let $\preceq$ be a total order on $H$ for which $(H, \preceq)$ is a linearly ordered monoid, that is, $xz \prec yz$ and $zx \prec zy$ for all $x, y, z \in H$ with $x \prec y$ (as usual, $u \prec v$ means $u \preceq v$ and $u \ne v$). Moreover, let $\delta$ be the canonical embedding $H \to R[H]$, so that $\delta(x)$ is, for every $x \in H$, a Kronecker delta $H \to R$ centered at $x$, which I'll rather denote by $\delta_x$.

Now pick $x \in H$, and assume $\delta_x = f\ast g$ for some $f, g \in R[H]$, with $\ast$ being the multiplication in $R[H]$. I claim that $|{\rm supp}(f)| = |{\rm supp}(g)| = 1$. Indeed, it is clear that $f$ and $g$ cannot be identically zero, so the support of each of them is non-empty. Accordingly, let $y_f$ and $z_f$ be, respectively, the minimum and the maximum of $S_f := {\rm supp}(f)$ relative to the order $\preceq$ (which exist because $\preceq$ is total and $S_f$ is not only non-empty, but also finite); define $y_g$ and $z_g$ in a similar way (only with $g$ in lieu of $f$). Then set $y := y_f y_g$ and $z := z_f z_g$. We have $$ \delta_x(y) = \sum_{uv=y} f(u) g(v) = f(y_f) g(y_g) \ne 0_R.$$ To see this, note that, if $u \prec y_f$ or $v \preceq y_g$, then $f(u) g(v) = 0_R$, and on the other hand, if $y_f \preceq u$ and $y_g \preceq v$, then $y = y_f y_g \preceq uv$, with equality iff $u = y_f$ and $v = y_g$ (by the assumption that $H$ is linearly ordered by $\preceq$). Then the rest is trivial, because $y_f \in {\rm supp}(f)$ and $y_g \in {\rm supp}(g)$ give $f(y_f) g(y_g) \ne 0_R$ (by the fact that $R$ is a domain).

To wit, we have shown that $y$ is in the support of $\delta_x$, and so is $z$ by an analogous argument. It follows $x = y = z$, which is, however, possible only if $y_f = z_f$ and $y_g = z_g$ (by construction, $y_f \preceq z_f$ and $y_g \preceq z_g$, and we can't have $y_f \neq z_f$ or $y_g \ne z_g$, otherwise $H$ being linearly ordered by $\preceq$ would yield a contradiction).

In other words, $f = a\delta_u$ and $g = b \delta_v$ for some $a, b \in R$ and $u, v \in H$ such that $ab= 1_R$. But $R^\times$ is trivial (by hypothesis), and therefore $a = b = 1_R$. So $f, g \in \delta(H)$, and we are done. []

Figured it out (sorry for answering my own question). I'll prove the following:

Lemma. Let $H$ be a linearly orderable monoid and $R$ a domain whose group of units is trivial (e.g., $R$ could be the ring of integers mod $2$). Then $H$ embeds as a divisor-closed submonoid into the multiplicative monoid of the monoid ring $R[H]$.

This will provide an affirmative answer to the question in the OP, since $\mathscr B(G)$ is a torsion-free, cancellative, commutative monoid, and hence is linearly orderable.

Proof of the lemma. Let $\preceq$ be a total order on $H$ for which $(H, \preceq)$ is a linearly ordered monoid, that is, $xz \prec yz$ and $zx \prec zy$ for all $x, y, z \in H$ with $x \prec y$ (as usual, $u \prec v$ means $u \preceq v$ and $u \ne v$). Moreover, let $\delta$ be the canonical embedding $H \to R[H]$, so that $\delta(x)$ is, for every $x \in H$, a Kronecker delta $H \to R$ centered at $x$, which I'll rather denote by $\delta_x$.

Now pick $x \in H$, and assume $\delta_x = f\ast g$ for some $f, g \in R[H]$, with $\ast$ being the multiplication in $R[H]$. I claim that $|{\rm supp}(f)| = |{\rm supp}(g)| = 1$. Indeed, it is clear that $f$ and $g$ cannot be identically zero, so the support of each of them is non-empty. Accordingly, let $y_f$ and $z_f$ be, respectively, the minimum and the maximum of $S_f := {\rm supp}(f)$ relative to the order $\preceq$ (which exist because $\preceq$ is total and $S_f$ is not only non-empty, but also finite); define $y_g$ and $z_g$ in a similar way (only with $g$ in lieu of $f$). Then set $y := y_f y_g$ and $z := z_f z_g$. We have $$ \delta_x(y) = \sum_{uv=y} f(u) g(v) = f(y_f) g(y_g) \ne 0_R.$$ To see this, note that, if $u \prec y_f$ or $v \preceq y_g$, then $f(u) g(v) = 0_R$, and on the other hand, if $y_f \preceq u$ and $y_g \preceq v$, then $y = y_f y_g \preceq uv$, with equality iff $u = y_f$ and $v = y_g$ (by the assumption that $H$ is linearly ordered by $\preceq$). Then the rest is trivial, because $y_f \in {\rm supp}(f)$ and $y_g \in {\rm supp}(g)$ give $f(y_f) g(y_g) \ne 0_R$ (by the fact that $R$ is a domain).

To wit, we have shown that $y$ is in the support of $\delta_x$, and so is $z$ by an analogous argument. It follows $x = y = z$, which is, however, possible only if $y_f = z_f$ and $y_g = z_g$ (by construction, $y_f \preceq z_f$ and $y_g \preceq z_g$, and we can't have $y_f \neq z_f$ or $y_g \ne z_g$, otherwise $H$ being linearly ordered by $\preceq$ would yield a contradiction).

In other words, $f = a\delta_u$ and $g = b \delta_v$ for some $a, b \in R$ and $u, v \in H$ such that $ab= 1_R$. But $R^\times$ is trivial (by hypothesis), and therefore $a = b = 1_R$. So $f, g \in \delta(H)$, and we are done. []