I am grateful to Edgardo whose answer supplied me with the Siegel theorem on integral points (I did not know it), and I am happy to see that my approach coincides with Edgardo's in many points.

The answer is negative: there is no abelian subgroup $A\subset{\mathbb R}^2$ of rank $3$ such that its intersection $I_A:={\mathbb S}^1\cap A$ with the unit circle ${\mathbb S}^1:=\big\{v\in{\mathbb R}^2\mid|v|=1\big\}$ is infinite.

In a few words: we show that such $I_A$ is a part of the intersection of two quadrics in ${\mathbb P}_{\mathbb C}^3$ defined over ${\mathbb Q}$, observe
that this intersection in interesting cases is generic, hence, is an elliptic curve, and, finally, apply the Siegel theorem http://en.wikipedia.org/wiki/Siegel_theorem to an affine part of this curve.

Let $L_{\mathbb Q}:={\mathbb Q}A$ denote the ${\mathbb Q}$-linear space spanned by $A$. Besides the intersection $I_A$, we will also deal with the intersection $I_{\mathbb Q}:={\mathbb S}^1\cap L_{\mathbb Q}$. One can choose another ${\mathbb Q}$-basis $B'$ in $L_{\mathbb Q}$ and generate another $A'$. It is immediate that $A\subset\frac1nA'$ and $A'\subset\frac1nA$ for a suitable $n\in{\mathbb N}$. We conclude that $I_A$ is infinite for some $A$ iff, for any ${\mathbb Q}$-basis $B$ in $L_{\mathbb Q}$, there are infinitely many elements in $I_{\mathbb Q}$ whose denominators in terms of $B$ are limited.

Assuming that some $I_A$ is infinite, we can choose a ${\mathbb Q}$-basis of $L_{\mathbb Q}$ inside $I_{\mathbb Q}$. Indeed, it is easy to pick a couple of ${\mathbb R}$-linearly independent $b_1,b_2\in I_{\mathbb Q}$. They generate a discrete ${\mathbb Z}$-lattice $A_0\subset{\mathbb R}^2$; so, $\frac1nA_0$ is also discrete for any $n\in{\mathbb N}$. Consequently, the intersection ${\mathbb S}^1\cap\frac1nA_0$ is finite. Including $b_1,b_2$ in some basis of $L_{\mathbb Q}$, we can see that $I_{\mathbb Q}\subset{\mathbb Q}A_0$ would imply $I_A\subset\frac1nA_0$ for a suitable $n\in{\mathbb N}$. A contradiction. For a similar reason, any $2$-dimensional ${\mathbb Q}$-linear subspace in $L_{\mathbb Q}$ contains finitely many rational points from $I_{\mathbb Q}$ with limited denominators.

Let $b_1,b_2,b_3\in I_{\mathbb Q}$ be a ${\mathbb Q}$-basis in $L_{\mathbb Q}$. Its Gram matrix has the form $G:=\left[\begin{smallmatrix}1&c_3&c_2\\c_3&1&c_1\\c_2&c_1&1\end{smallmatrix}
\right]$
with $-1<c_i<1$. Since the $b_i$'s are ${\mathbb R}$-linearly dependent,
$\det G=0$. So, if $c_1,c_2,c_3\in{\mathbb Q}$, then, by means of the Gram-Schmidt orthogonalization in $L_{\mathbb Q}$ and by the Sylvester criterion, we could find, in view of $\det G=0$, an isotropic element
$0\ne v\in L_{\mathbb Q}$, i.e., such that $\langle v,v\rangle=0$, which is impossible. Hence, $c_i\notin{\mathbb Q}$ for some $i$.

Actually, we study rational solutions of
$$x_1^2+x_2^2+x_3^2+2c_1x_2x_3+2c_2x_3x_1+2c_3x_1x_2=1.\qquad(1)$$

As $c_3^2<1$, there are finitely many rational solutions of the form $[x_1,x_2,0]$ with limited denominators, because $(1)$ becomes
$(x_1+c_3x_2)^2+(1-c_3^2)x_2^2=1$. In other words, we can assume that $x_1x_2x_3\ne0$ and rewrite $(1)$ in terms of $y_i:=x_i^{-1}$ as
$$\frac{y_1y_2}{y_3}+\frac{y_2y_3}{y_1}+\frac{y_3y_1}{y_2}+2c_1y_1+2c_2y_2+2c_3y_3=y_1y_2y_3.\qquad(2)$$

The dimension of the ${\mathbb Q}$-linear space spanned by $[x_2x_3,x_3x_1,x_1x_2]$, where $[x_1,x_1,x_3]$ runs over all rational solutions of $(1)$, equals $2$. Indeed, it cannot be $3$ as, otherwise, $c_1,c_2,c_3\in{\mathbb Q}$. It cannot be $0$ because we have at least one rational solution. If the dimension equals $1$, then, for all rational solutions with $x_1x_2x_3\ne0$, the triples $[y_1,y_2,y_3]$ are all ${\mathbb Q}$-proportional. Looking at $(2)$, we see that the coefficient of proportionality should be $\pm1$, and we get only finitely many solutions of $(1)$ with limited denominators.

We conclude that $c_1=a_1c+d_1$, $c_2=a_2c+d_2$, $c_3=c$ for suitable $a_1,a_2,d_1,d_2\in{\mathbb Q}$. It follows from $\det G=0$ that $p(c)=0$, where $p(x):=\det\left[\begin{smallmatrix}1&x&d_2+a_2x\\x&1&d_1+a_1x\\d_2+a_2x&d_1+a_1x&1\end{smallmatrix}\right]$.

Now, $(1)$ takes the form
$x_1^2+x_2^2+x_3^2+2d_1x_2x_3+2d_2x_3x_1+2(x_1x_2+a_1x_2x_3+a_2x_3x_1)c=1$. Since $c\notin{\mathbb Q}$, we deal in fact with the equations
$$x_1^2+x_2^2+x_3^2+2d_1x_2x_3+2d_2x_3x_1=1,\quad x_1x_2+a_1x_2x_3+a_2x_3x_1=0.$$

We want to show that, for any integer $0\ne d\in{\mathbb Z}$, the equations
$$x_1^2+x_2^2+x_3^2+2d_1x_2x_3+2d_2x_3x_1=d^2,\quad2x_1x_2+2a_1x_2x_3+2a_2x_3x_1=0$$
admit just finitely many integer solution.

The intersection
$$x_1^2+x_2^2+x_3^2+2d_1x_2x_3+d_2x_3x_1-d^2x_0^2=0,\quad2x_1x_2+2a_1x_2x_3+2a_2x_3x_1=0$$
of two quadrics in ${\mathbb P}_{\mathbb C}^3$ is an elliptic curve if all the $4$ roots of the polynomial
$q(\lambda):=\det(\lambda M_1+M_2)$ are pairwise distinct, where $M_1,M_2$ stand for the symmetric matrices of the quadrics (see, for example,
http://archive.numdam.org/ARCHIVE/ASNSP/ASNSP_1980_4_7_2/ASNSP_1980_4_7_2_217_0/ASNSP_1980_4_7_2_217_0.pdf
especially, from pages 221-222). Since
$q(\lambda)=\det\left[\begin{smallmatrix}-d^2\lambda&0&0&0\\0&\lambda&1&d_2
\lambda+a_2\\0&1&\lambda&d_1\lambda+a_1\\0&d_2\lambda+a_2&d_1\lambda+
a_1&\lambda\end{smallmatrix}\right]=-d^2\lambda^4p(\lambda^{-1})$,
in view of the Siegel theorem, it suffices to show that all the $3$ roots of the polynomial $\lambda^3p(\lambda^{-1})=0$ are $(\text{a})$ pairwise distinct and $(\text{b})$ all different from $0$. The polynomial $p(x)$ of degree $\le3$ has rational coefficients, $p(c)=0$, and $c\notin{\mathbb Q}$. Consequently, $p(x)$ cannot have multiple roots thus implying $(\text{a})$. If $(\text{b})$ is not true, the coefficient of degree $3$ in $p(x)$ vanishes, i.e., $2a_1a_2=0$. In this case, the second quadric is a couple of rational planes passing through the origin $0\in L_{\mathbb Q}$, and we already know that any $2$-dimensional ${\mathbb Q}$-linear subspace in $L_{\mathbb Q}$ contains finitely many rational points from $I_{\mathbb Q}$ with limited denominators.