Yes: a f.g. discrete, and more generally compactly generated locally compact group is doubling iff it has polynomial growth.

For f.g. groups, you mentioned $\Rightarrow$, and asked $\Leftarrow$, which I justify below.

Define $X$ to be large-scale doubling if for some $R_0,M_0$, every ball of radius $R\ge R_0$ is finite union of $M_0$ balls of radius $R/2$.

To be large-scale doubling is a QI-invariant. For a metric space in which balls of given radius have bounded cardinal, it's obviously equivalent to doubling.

So for a f.g. group, being doubling doesn't depend on a choice of finite generating subset. Since every f.g. nilpotent group is QI to some simply connected nilpotent Lie group (Malcev), it is enough to check that every simply connected nilpotent Lie group $G$ is large-scale doubling. (More generally every compactly generated locally compact group of at most polynomial growth is QI to such a $G$.)

Indeed Pansu proved in 1983 that every asymptotic cone of such a Lie group $G$ is homeomorphic to $G$ and is a proper metric space. This prevents $G$ to be large-scale doubling, by the following fact:

If a space $X$ is not large-scale doubling, then there exists a sequence of points $(x_n)$, and radii $r_n\to\infty$ and $M_n\to\infty$ such that the $2r_n$-ball around $x_n$ contains $M_n$ points at distance $\ge r_n$. It easily follows that the ultralimit of rescaled metric spaces $(X,x_n,\frac{1}{r_n}d)$, which has a natural basepoint $o$ has infinitely many points in the $2$-ball around $o$ at pairwise distance $\ge 1$, so is not a proper metric space.

Yes: a f.g. discrete, and more generally compactly generated locally compact group is doubling iff it has polynomial growth.

For f.g. groups, you mentioned $\Rightarrow$, and asked $\Leftarrow$, which I justify below.

Define $X$ to be large-scale doubling if for some $R_0,M_0$, every ball of radius $R\ge R_0$ is finite union of $M_0$ balls of radius $R/2$.

To be large-scale doubling is a QI-invariant. For a metric space in which balls of given radius have bounded cardinal, it's obviously equivalent to doubling.

So for a f.g. group, being doubling doesn't depend on a choice of finite generating subset. Since every f.g. nilpotent group is QI to some simply connected nilpotent Lie group (Malcev), it is enough to check that every simply connected nilpotent Lie group $G$ is large-scale doubling. (More generally every compactly generated locally compact group of at most polynomial growth is QI to such a $G$.)

Indeed Pansu proved in 1983 that every asymptotic cone of such a Lie group $G$ is homeomorphic to $G$ and is a proper metric space. This implies that $G$ is large-scale doubling, by the following fact:

If a space $X$ is not large-scale doubling, then there exists a sequence of points $(x_n)$, and radii $r_n\to\infty$ and $M_n\to\infty$ such that the $2r_n$-ball around $x_n$ contains $M_n$ points at distance $\ge r_n$. It easily follows that the ultralimit of rescaled metric spaces $(X,x_n,\frac{1}{r_n}d)$, which has a natural basepoint $o$ has infinitely many points in the $2$-ball around $o$ at pairwise distance $\ge 1$, so is not a proper metric space.

Yes: a f.g. discrete, and more generally compactly generated locally compact group is doubling iff it has polynomial growth.

For f.g. groups, you mentioned $\Rightarrow$, and asked $\Leftarrow$, which I justify below.

Define $X$ to be large-scale doubling if for some $R_0,M_0$, every ball of radius $R\ge R_0$ is finite union of $M_0$ balls of radius $R/2$.

To be large-scale doubling is a QI-invariant. For a metric space in which balls of given radius have bounded cardinal, it's obviously equivalent to doubling.

So for a f.g. group, being doubling doesn't depend on a choice of finite generating subset. Since every f.g. nilpotent group is QI to some simply connected nilpotent Lie group (Malcev), it is enough to check that every simply connected nilpotent Lie group $G$ is large-scale doubling. (More generally every compactly generated locally compact group of at most polynomial growth is QI to such a $G$.)

Indeed Pansu proved in 1983 that every asymptotic cone of such a group is homeomorphic to $G$ and is a proper metric space. This prevents $G$ to be large-scale doubling, by the following fact:

If a space $X$ is not large-scale doubling, then there exists a sequence of points $(x_n)$, and radii $r_n\to\infty$ and $M_n\to\infty$ such that the $2r_n$-ball around $x_n$ contains $M_n$ points at distance $\ge r_n$. It easily follows that the ultralimit of rescaled metric spaces $(X,x_n,\frac{1}{r_n}d)$, which has a natural basepoint $o$ has infinitely many points in the $2$-ball around $o$ at pairwise distance $\ge 1$, so is not a proper metric space.

Yes: a f.g. discrete, and more generally compactly generated locally compact group is doubling iff it has polynomial growth.

For f.g. groups, you mentioned $\Rightarrow$, and asked $\Leftarrow$, which I justify below.

Define $X$ to be large-scale doubling if for some $R_0,M_0$, every ball of radius $R\ge R_0$ is finite union of $M_0$ balls of radius $R/2$.

To be large-scale doubling is a QI-invariant. For a metric space in which balls of given radius have bounded cardinal, it's obviously equivalent to doubling.

So for a f.g. group, being doubling doesn't depend on a choice of finite generating subset. Since every f.g. nilpotent group is QI to some simply connected nilpotent Lie group (Malcev), it is enough to check that every simply connected nilpotent Lie group $G$ is large-scale doubling. (More generally every compactly generated locally compact group of at most polynomial growth is QI to such a $G$.)

Indeed Pansu proved in 1983 that every asymptotic cone of such a Lie group $G$ is homeomorphic to $G$ and is a proper metric space. This prevents $G$ to be large-scale doubling, by the following fact:

If a space $X$ is not large-scale doubling, then there exists a sequence of points $(x_n)$, and radii $r_n\to\infty$ and $M_n\to\infty$ such that the $2r_n$-ball around $x_n$ contains $M_n$ points at distance $\ge r_n$. It easily follows that the ultralimit of rescaled metric spaces $(X,x_n,\frac{1}{r_n}d)$, which has a natural basepoint $o$ has infinitely many points in the $2$-ball around $o$ at pairwise distance $\ge 1$, so is not a proper metric space.