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Yes, the set of points attainable by these paths is dense in $\mathbb{R}^2$. Pardon the lack of polish in this answer; I don't have a 'real' math background, so I'm expressing my ideas here using the language most intuitive to me.

We can think of any path as a linear combination of five 'moves,' which w.l.o.g. are given by the vectors $v_n = (\cos\frac{2\pi n}{5}, \sin \frac{2\pi n}{5})$. Paths cannot be arbitrary integer combinations of these vectors, since there are only two moves available after each step. We can, however, devise the following scheme to obtain arbitrary integer combinations of the vectors $2v_n$. Consider the following paths (I've written out the linear combination long form to show the order that the steps must be taken in each path):

\begin{align*} p_1^a &= 2av_1\\ p_2^b &= v_1 + 2bv_2 + v_1\\ p_3^c &= v_1 + v_2 + 2cv_3 + v_2 + v_1\\ p_4^d &= v_1 + v_2 + v_3 + 2dv_4 + v_3 + v_2 + v_1\\ p_5^e &= 2ev_5 \end{align*}

Graphically, these look something like this: $p_1^2$, $p_2^2$, $p_3^2$, $p_4^2$, $p_5^2$.

These paths are composable with each other, so we can take them in arbitrary (positive) integer combinations. Consider a point of the form $x = \sum_{i = 1}^5 p_i^{a_i}$. Writing this out in terms of our original moves $v_i$, we obtain \begin{align*} x &= p_1^a + p_2^b + p_3^c + p_4^d + p_5^e\\ &= (6 + 2a)v_1 + (4 + 2b)v_2 + (2 + 2c)v_3 + 2dv_4 + 2ev_5 \end{align*} We will ignore the constant offset $6v_1 + 4v_2 + 2v_3$. Using the paths $p_i^a$ alone, we can thus reach an arbitrary (nonnegative) integer combination of the vectors $\{2v_i\}$.

Now note that $\sum_{i = 1}^5 2v_i = 0$, so it follows that the vectors $2v_i$ and $\sum_{j

eq i} 2v_j$ are additive inverses. Thus the set of nonnegative integer combinations of the vectors $2v_i$ is actually equal to the set of arbitrary integer combinations of the vectors $2v_i$.

Digging around MathSE, I found this answer that asserts, based on Kronecker's theorem, that if three vectors have components that are all linearly independent over the rationals, then the set of their integer combinations is dense in the unit square (and by extension the plane). $2v_1$, $2v_2$ and $2v_3$ satisfy this hypothesis, so the claim should follow.