For $b=1$, the turtle can only reach equidistant points within a ray;
it cannot move backwards, so that case does not yield a group.
In the following I will assume that $b$ is an integer greater than $1$.
Let the turtle start at $0$ in the complex plane with direction $1$.
There are exactly $b$ possible directions for the turtle,
and we can represent those directions as
$\zeta_b^k = \exp(\mathrm{i}k\delta)
= \exp\left(\frac{2\pi\mathrm{i}k}{b}\right)$
for $k\in\{0,\ldots,b-1\}$.
For each direction $\zeta_b^k$ we can maintain a step counter
$a_k\in\mathbb{N}_0$ that is incremented whenever the turtle steps
in that direction.
Then the turtle's position can be given as
$$z = \sum_{k=0}^{b-1} a_k\zeta_b^k$$
I claim that the set of all possible $z$ remains unchanged
if we allow negative integer values for the $a_k$ as well.
In other words, the equivalent of backward moves is possible.
This can be deduced from the fact that
$$\sum_{k=0}^{b-1} \zeta_b^k = 0\quad\text{for $b>1$}$$
(Note that this requires $b>1$ which is why that case had to be separated.)
Concretely, let $a_{\text{min}} = \min_{k=0,\ldots,b-1} a_k$, then
$$\sum_{k=0}^{b-1} \underbrace{(a_k - a_{\text{min}})}_{\geq0}\,\zeta_b^k
= \underbrace{\left(\sum_{k=0}^{b-1} a_k\zeta_b^k\right)}_z
- a_{\text{min}}\underbrace{\sum_{k=0}^{b-1} \zeta_b^k}_0 = z$$
So we can make all step counters nonnegative by adding $-a_{\text{min}}$
to all of them without changing the turtle position $z$.
As a result, the set of all possible turtle positions can be identified
with $\mathbb{Z}[\zeta_b]$, the ring of all polynomials in $\zeta_b$
with integer coefficients.
Another thread
unveils the cases when $\mathbb{Z}[\zeta_b]$ is dense in $\mathbb{C}$:
Since $\zeta_b$ is an algebraic integer (it solves $X^b - 1 = 0$),
$\mathbb{Z}[\zeta_b]$ is dense in $\mathbb{C}$ if and only if
the degree of (the minimal polynomial of) $\zeta_b$ over $\mathbb{Q}$
is greater than $2$.
The minimal polynomial of $\zeta_b$ over $\mathbb{Q}$ is the
cyclotomic polynomial
$\Phi_b$, its degree is $\phi(b)$ where $\phi$ is Euler's totient function:
$$\phi(b) = b\prod_{\substack{p\text{ prime}\\p\mid b}}\frac{p-1}{p}$$
$\phi(b)$ is a positive integer,
so we just need to exclude those $b$ that have $\phi(b)\in\{1,2\}$.
These are $b \in \{1,2,3,4,6\}$; to see that there are no more $b$,
note that
- if $b$ had a prime divisor $p$ greater than $3$, then
$\phi(b)$ would be divisible by $p-1$ which is greater than $2$;
- if $b$ were divisible by $2^3$ or $2^2\cdot3$,
then $\phi(b)$ would be divisible by $4$; and
- if $b$ were divisible by $3^2$,
then $\phi(b)$ would be divisible by $6$.
Consequently, $\mathbb{Z}[\zeta_b]$ is dense in $\mathbb{C}$ if and only if
$b\not\in\{1,2,3,4,6\}$.
@Rahul has indicated another approach which may seem easier to grasp:
Obviously, $b>2$ is necessary and sufficient for the turtle to move in more than
one dimension. So let us assume $b>2$.
Note that $+F--F+$ moves forward $2\cos\delta$.
Replacing $F$ with $(+F)^{b-1}+$ moves backward instead.
So we can effectively move in the current direction with any step size
from the $\mathbb{Z}$-span of $1$ and $2\cos\delta$.
Therefore, if $\cos\delta\not\in\mathbb{Q}$, the set of available
step sizes is dense in $\mathbb{R}$. Combined with the ability to move in
more than one dimension once $b>2$, this makes the set of reachable turtle
positions dense in $\mathbb{R}^2$ if $\cos\delta$ is irrational.
In yet another thread,
a useful lemma has been given:
If $\frac{\delta}{\pi}$ is rational, then $2\cos\delta$ is an algebraic integer.
Proof: Assuming $\frac{3\delta}{\pi}=\frac{c}{n}$
with $c,n$ coprime integers and $n>0$, we have
$$2\cos(n\delta)=2\cos\frac{c\pi}{3}=u
\quad\text{for some}\quad u\in\{\pm1,\pm2\}$$ and
$$2\cos(n\delta) = D_n(x)\quad\text{where}\quad x = 2\cos\delta\tag{*}$$
where the $D_n(x)$ are Dickson polynomials of the first kind
(with second parameter $\alpha=1$).
Their essential property is
$$D_n\left(z+\frac{1}{z}\right) = z^n + \frac{1}{z^n}$$
Setting $z=\exp(\mathrm{i}\delta)$ yields $(*)$.
Those Dickson polynomials can be defined recursively as
$$\begin{align}
D_0(x) &= 2 \\
D_1(x) &= x \\
D_n(x) &= x\,D_{n-1}(x) - D_{n-2}(x) &&\text{for $n>1$}
\end{align}$$
By induction we can see that the Dickson polynomial $D_n$ has integral
coefficients, degree $n$ and is monic for $n>0$.
The same holds for $D_n(x) - u$ (at least for $n>0$ which we assume),
a root of which is $x=2\cos\delta$.
Therefore $2\cos\delta$ is an algebraic integer.
Consequently,
if $\cos\delta$ is rational, then $2\cos\delta$ must be an integer.
Since $\cos\delta$ is bounded to not exceed absolute value $1$,
the only possibilities are $2\cos\delta\in\{-2,-1,0,1,2\}$
which correspond to $b\in\{2,3,4,6,1\}$.
All other $b$ therefore yield irrational $\cos\delta$
and result in turtle position sets that are dense in $\mathbb{R}^2$.
