Quite remarkable

$\begin{aligned} xy + x + y = 23 \Rightarrow xy + x + y + 1 & = (x + 1)(y + 1) = 24 \\yz + y + z = 31 \Rightarrow yz + y + z + 1 & = (y + 1)(z + 1) = 32 \\zx + z + x = 47 \Rightarrow zx + z + x + 1 & = (z + 1)(x + 1) = 48\end{aligned}$

$\Rightarrow (x + 1)^2\times{(y + 1)^2}\times{(z + 1)^2} = 24\times32\times48 = 192^2$

$\Rightarrow (x + 1)\times{(y + 1)}\times{(z + 1)} = 192$ or $(x + 1)\times{(y + 1)}\times{(z + 1)} = -192$ .

The former case leads to $(z + 1) = \dfrac{192}{24} = 8 , (x + 1) = \dfrac{192}{32} = 6 , (y + 1) = \dfrac{192}{48} = 4 \Rightarrow \boxed{(x , y ,z) = (5 , 3 , 7)}$ .

The latter case leads to $(z + 1) = \dfrac{-192}{24} = -8 , (x + 1) = \dfrac{-192}{32} = -6 , (y + 1) = \dfrac{-192}{48} = -4 \Rightarrow \boxed{(x , y ,z) = (-7 , -5 , -9)}$ .

$xy+x+y = 23 \implies$ Eq.(1)

$yz+y+z = 31\\ z(y+1) = 31 - y$

$z=\dfrac{31-y}{y+1} \implies$ Eq.(2)

$xz+x+z = 47\\ z(x+1) = 47-x$

$z=\dfrac{47-x}{x+1} \implies$ Eq.(3)

Given that Eq.(2) = Eq.(3)

$\dfrac{31-y}{y+1} = \dfrac{47-x}{x+1}\\ (31-y)(x+1) = (47-x)(y+1)\\ 31x+31-xy-y = 47y+47-xy-x\\ 32x=48y + 16\\ 2x=3y+1\\ 3y=2x-1$

$y=\dfrac{2x-1}{3} \implies$ Eq.(4)

Substitute Eq.(4) into Eq.(1):

$x\left(\dfrac{2x-1}{3}\right) + x + \dfrac{2x-1}{3} = 23\\ x(2x-1) + 3x + (2x-1) = 69\\ 2x^2 - x + 3x + 2x -1 -69 = 0\\ 2x^2 + 4x - 70 = 0\\ x^2 + 2x - 35 = 0\\ (x + 7)(x - 5) = 0\\ x=-7,\;5$

When $x=-7$ :

$y=\dfrac{2(-7)-1}{3} = \dfrac{-15}{3} = -5\\ z=\dfrac{47-(-7)}{-7+1} = \dfrac{54}{-6} = -9$

When $x=5$ :

$y = \dfrac{2(5) - 1}{3} = \dfrac{9}{3} = 3\\ z=\dfrac{47-5}{5+1} = \dfrac{42}{6} = 7$

Sum of all possible values of $x$ , $y$ and $z$

$=-7-5-9+5+3+7 = \boxed{-6}$