When $x \ge 0$, $y = 1$; when $x < 0$, $y = \frac{x+1}{1-x}$ is strictly increasing about $x$ and less than $1$.

Suppose line $l: y = kx + b$, and then the known conditions are equivalent to the fact that equation
$$kx + b = \frac{x+1}{|x|+1} \qquad (1)$$
has three different real number solutions $x_1, x_2, x_3$ ($x_1 < x_2 < x_3$) satisfying $x_1 + x_2 + x_3 = 0$.

Firstly, there is $k \neq 0$, otherwise the function of $l$ can only be $y = 1$, but then the sum of the abscissas of any three common points of the graph of $l$ and function $y = \frac{x+1}{|x|+1}$ must be greater than $0$. This is not consistent with the question.

When $x < 0$, Equation (1) can be arranged as
$$kx^2 - (k - b - 1)x + 1 - b = 0, \qquad (2)$$
and it has at most two negative solutions.

When $x \ge 0$, Equation (1) follows as
$$kx + b = 1, \qquad (3)$$
and it has at most one non-negative solution.

This shows that Equation (2) has two different negative solutions $x_1, x_2$, where
$$x_1 + x_2 = \frac{k - b - 1}{2};$$
Equation (3) has a non-negative solution $x_3 = \frac{1-b}{k}$. By $x_1 + x_2 + x_3 = 0$, we know that $k = 2b$.

And then there is $x_3 = \frac{1-b}{2b}$. Since $x_3 \ge 0$, we have $0 < b \le 1$.

Equation (2) becomes $2b x^2 + (1-b)x + 1 - b = 0$. By discriminant
$$(1-b)^2 - 4 \cdot 2b(1-b) = (1-b)(1-9b) > 0$$
and combining $0 < b \le 1$, we have $0 < b < \frac{1}{9}$. (After checking, $x_1, x_2$ are indeed negative at this point, which is consistent with the question.)

In conclusion, the range of the slope $k = 2b$ of $l$ is $0 < k < \frac{2}{9}$.