In this question, it is shown that $(\mathbb{R}, +)$ is not a free group. But my question is: if it is not a free group, exactly what relations is it subject to?
My other question is: are there sets other than $\mathbb{R}$ that generate $(\mathbb{R}, +)$?
On
$(\mathbb{R}, +)$ is* a $\mathbb{Q}$-vector space whose dimension is the cardinality of $|\mathbb{R}|$. So the main things you need to think about to answer whatever questions you have are:
*: The vector space structure is uniquely determined from the group structure.
**: This invokes the axiom of choice, I think
To see that $(\mathbb{R},+)$ is not a free (abelian) group, note that $\mathbb{R}$ is divisible: given any $r\in\mathbb{R}$ and any $n\gt 0$, there exists $s\in\mathbb{R}$ such that $ns = r$.
But a free (abelian) group is never divisible: if $F$ is a free abelian group, and $x$ is a basis element, then there is no element of $F$ such that $2y=x$. If there were, we would be able to write $y = k_1x_1+\cdots + k_mx_m$ for some distinct basis elements $x_1,\ldots,x_m$ and integers $k_i$. Then $x = 2k_1x_1+\cdots 2k_mx_m$, so then there is at most one nonzero $k_i$, say $k_1$, $x_1=x$, and $2k_1 = 1$, which is impossible. Thus, free abelian groups cannot be divisible, and therefore free groups cannot be divisible (since quotients of divisible groups are divisible, and a free abelian group is a quotient of a free group).
So $\mathbb{R}$ is subject to infinitely many relations, some of which tell you any two elements commute (it satisfies the identity $x+y-x-y$); others that tell you that certain elements are "half" other elements, and so on.
Explicitly producing generating sets is not hard: any cofinite set generates $\mathbb{R}$ as an abelian group; the set of reals on $[0,1)$ generate $\mathbb{R}$.