One can use your infinite-density example, but replace the outer lines with very sparse dotted lines:

$$S = (\{0\} \times \mathbb{R}) \cup \bigcup_{i=1}^\infty \{i^{-1},-i^{-1}\} \times \left[\bigcup_{j \in \mathbb{Z}} [i^{-2}j,i^{-2}j+i^{-4}]\right] $$

By symmetry it is enough to consider the intersection with $[0,1]^2$ for local finiteness. There, what is left of the $i$-th line consists of $i^2$ pieces of length $i^{-4}$, i.e. contributes a total of $i^{-2}$ to the Hausdorff measure, so their sum is finite.

Additionally, each of the line-pieces is closed and isolated, their only accumulation points are on the line $\{0\}\times \mathbb{R}$, which is already included in $S$, so $S$ is closed.

Finally, take $x=(0,x_2)$ and $v=(v_1,v_2) \in \mathbb{R}^2$. For $v_1=0$ it is clear that $v$ is in the tangent cone. For $v_1 \neq 0$ and any $i \in \mathbb{N}$ pick $r = \frac{1}{i|v_1|}$. Then $x+rv = (\frac{\pm1}{i},x_2+rv_2)$, which no matter the second component is of distance less than $i^{-2}=r^2|v_1|^2$ to a point of the form $(\pm i^{-1},i^{-2}j)$ and thus in $S$. This gives you a sequence for the liminf and shows that $v$ is in the tangent cone of $x$.

Also as an additional sanity check, note that this does not violate the Besicovitch-Federer structure theorem: While the tangent cones at $x\in {0}\times \mathbb{R}$ are degenerate and the set has positive integralgeometric measure, we have that due to the low density of $S$, the approximate tangent cone at each $x\in {0}\times \mathbb{R}$ is still ${0}\times \mathbb{R}$.

One can use your infinite-density example, but replace the outer lines with very sparse dotted lines:

$$S = (\{0\} \times \mathbb{R}) \cup \bigcup_{i=1}^\infty \{i^{-1},-i^{-1}\} \times \left[\bigcup_{j \in \mathbb{Z}} [i^{-2}j,i^{-2}j+i^{-4}]\right] $$

By symmetry it is enough to consider the intersection with $[0,1]^2$ for local finiteness. There, what is left of the $i$-th line consists of $i^2$ pieces of length $i^{-4}$, i.e. contributes a total of $i^{-2}$ to the Hausdorff measure, so their sum is finite.

Additionally, each of the line-pieces is closed and isolated, their only accumulation points are on the line $\{0\}\times \mathbb{R}$, which is already included in $S$, so $S$ is closed.

Finally, take $x=(0,x_2)$ and $v=(v_1,v_2) \in \mathbb{R}^2$. For $v_1=0$ it is clear that $v$ is in the tangent cone. For $v_1 \neq 0$ and any $i \in \mathbb{N}$ pick $r = \frac{1}{i|v_1|}$. Then $x+rv = (\frac{\pm1}{i},x_2+rv_2)$, which no matter the second component is of distance less than $i^{-2}=r^2|v_1|^2$ to a point of the form $(\pm i^{-1},i^{-2}j)$ and thus in $S$. This gives you a sequence for the liminf and shows that $v$ is in the tangent cone of $x$.

Also as an additional sanity check, note that this does not violate the Besicovitch-Federer structure theorem: While the tangent cones at $x\in \{0\}\times \mathbb{R}$ are degenerate and the set has positive integralgeometric measure, we have that due to the low density of $S$, the approximate tangent cone at each $x\in \{0\}\times \mathbb{R}$ is still $\{0\}\times \mathbb{R}$.