Solution:
Let us first expand and simplify the equations.

From the first equation:
$$x(y-1) + y(x+1) = 6$$
Expand:
$$xy - x + yx + y = 6$$
Combine like terms:
$$xy - x + xy + y = 6$$
$$2xy - x + y = 6$$

From the second equation:
$$(x-1)(y+1) = 1$$
Expand:
$$xy + x - y - 1 = 1$$
$$xy + x - y = 2$$

Now, let us solve the system:

We have:
1) $2xy - x + y = 6$
2) $xy + x - y = 2$

Let us add the two equations:

Equation 1: $2xy - x + y = 6$
Equation 2: $xy + x - y = 2$

Let us try to eliminate variables. Add equation 1 and equation 2:

But first, let's try to express one variable in terms of the other.

From equation 2:
$$xy + x - y = 2$$
$$xy - y = 2 - x$$
$$y(x - 1) = 2 - x$$
If $x \neq 1$,
$$y = \frac{2 - x}{x - 1}$$

Now substitute this value of $y$ into equation 1:
$$2xy - x + y = 6$$
Substitute $y$:
$$2x \left(\frac{2 - x}{x - 1}\right) - x + \frac{2 - x}{x - 1} = 6$$
Multiply both sides by $x - 1$ (assuming $x \neq 1$):
$$2x(2 - x) - x(x - 1) + (2 - x) = 6(x - 1)$$
Expand:
$$2x(2 - x) = 4x - 2x^2$$
$-x(x - 1) = -x^2 + x$
So:
$$4x - 2x^2 - x^2 + x + 2 - x = 6x - 6$$
Combine like terms:
$-2x^2 - x^2 = -3x^2$
$4x + x - x = 4x$
So:
$$-3x^2 + 4x + 2 = 6x - 6$$
Bring all terms to one side:
$$-3x^2 + 4x + 2 - 6x + 6 = 0$$
$$-3x^2 - 2x + 8 = 0$$
Multiply both sides by $-1$:
$$3x^2 + 2x - 8 = 0$$
Now solve this quadratic equation:
$$x = \frac{-2 \pm \sqrt{4 + 96}}{6} = \frac{-2 \pm \sqrt{100}}{6} = \frac{-2 \pm 10}{6}$$
So:
$$x_1 = \frac{-2 + 10}{6} = \frac{8}{6} = \frac{4}{3}$$
$$x_2 = \frac{-2 - 10}{6} = \frac{-12}{6} = -2$$
Now, recall $y = \frac{2 - x}{x - 1}$.

For $x = \frac{4}{3}$:
$$y = \frac{2 - \frac{4}{3}}{\frac{4}{3} - 1} = \frac{\frac{6 - 4}{3}}{\frac{4 - 3}{3}} = \frac{\frac{2}{3}}{\frac{1}{3}} = 2$$
For $x = -2$:
$$y = \frac{2 - (-2)}{-2 - 1} = \frac{4}{-3} = -\frac{4}{3}$$

Therefore, the real solutions are:
$$(x, y) = \left(\frac{4}{3}, 2\right),\quad (x, y) = \left(-2, -\frac{4}{3}\right)$$

Check for $x = 1$:
From above, $y = \frac{2 - x}{x - 1}$ is undefined for $x = 1$, so no solution for $x = 1$.

Final Answer:
All real solutions are:
$$\boxed{\left(\frac{4}{3}, 2\right),\ \left(-2, -\frac{4}{3}\right)}$$