By changing $x, y$, it is obvious that the given set is the same as $\left\{\frac{f(x)-f(y)}{x-y} \mid x, y \in \mathbb{R}, x > y\right\}$. Let's define $F(x, y) = \frac{f(x)-f(y)}{x-y}$. First, we prove two lemmas:

Lemma. $0, 1 \notin I$. So $f(x)$ is an injective function.
Proof. Assume the contrary, so there is $x, y$ such that $F(x, y) = 1$, then $y + f(x) = x + f(y)$, so we find that $x = y$ which yields a contradiction. Now if $F(x, y) = 0$, for some $x, y$, then we find that $f(x) = f(y)$ which leads to
$$f(y + f(x)) - f(x + f(x)) = y - x$$
or $F(y + f(x), x + f(x)) = 1$, which is impossible.

Lemma. $F(x, y) > 1, \forall x \neq y$. So $f(x)$ is strictly increasing.
Proof. First, we rewrite the original equation as
$$f(x + f(y)) - f(y + f(y)) = f(y + f(x)) - f(y + f(y)) + x - y$$
Which, since $f$ is injective, is equivalent to
$$F(x + f(y), y + f(y)) = F(y + f(x), y + f(y)) \cdot F(x, y) + 1, \forall x \neq y.$$
Now since $0 \notin I$, we find that $F$ is either always positive or always negative, so the right-hand side of the above equality is strictly greater than 1 implying that $F(x + f(y), y + f(y)) > 1$. Since $1 \notin I$, we find that $F(x, y) > 1$ for all $x \neq y$.
Now, if $x < y$, given that $\frac{f(x)-f(y)}{x-y} > 1$, we have $f(x)+y < x+f(y)$, thus $f(f(x) + y) < f(x + f(y))$, resulting in $x > y$, a contradiction. So there is no such function. ■