Denote $x+1 = a$, $y+1 = b$ and $z+1 = c$. Adding 1 to the sides of all equations and factorizing in the left gives
$$\begin{cases} ab = c, \\ bc = a, \\ ca = b. \end{cases} \qquad (2)$$
If $a=0$ then the first and the third equation of (2) imply $b=0$ and $c=0$. Analogously, if $b=0$ or $c=0$ then $a, b, c$ are all equal to zero. This case leads to the solution $(-1,-1,-1)$ to the initial system. If none of $a, b, c$ is zero then pairwise multiplication of the equations of (2) and each time reducing similar factors from both sides gives $a^2 = b^2 = c^2 = 1$. Thus each of the numbers $a, b, c$ is either 1 or -1. The case $a=b=c=1$ satisfies (2) and leads to the solution $(0,0,0)$ of the initial system. The cases with exactly two of numbers $a, b, c$ being equal to -1 also satisfy (2) and lead to the solutions $(-2,-2,0)$, $(-2,0,-2)$ and $(0,-2,-2)$ of the initial system. Multiplying the corresponding sides of all equations of (2) and reducing by factor $abc$ gives $abc = 1$, whence there are no more solutions (the value -1 cannot occur an odd number of times).

Substituting $z$ from the first equation into the second one gives the equation $y(xy + x + y) + y + (xy + x + y) = x$ which is equivalent to $(x+1)y(y+2) = 0$. Thus we have $x = -1$ or $y = 0$ or $y = -2$.
* If $x = -1$ then the first equation implies $z = -1$. Substituting into the third equation gives $y = -1$. Hence the solution $(-1,-1,-1)$.

* If $y = 0$ then the first equation implies $z = x$. Substituting into the third equation gives $x^2 + 2x = 0$. Thus $x = z = 0$ or $x = z = -2$, leading to the solutions $(0, 0, 0)$ and $(-2, 0, -2)$.
* If $y = -2$ then the first equation implies $z = -x - 2$. Substituting it into the third equation gives $x^2 + 2x = 0$. Thus either $x = 0$ and $z = -2$ or $x = -2$ and $z = 0$ which lead to the solutions $(0, -2, -2)$ and $(-2, -2, 0)$.