Q:

Pick seven numbers so that every operation gives same result

Pick seven numbers so that every operation (addition, subtraction, multiplication, but not division) gives the same result.

For subtraction, you may take the result's absolute value.
Numbers cannot be $0$.

Example:  

$a, b, c, d, e, f, g$ are the numbers.  
$a + b + c + d + e + f + g = x$  
$| a - b - c - d - e - f - g |  = x$ or $ a - b - c - d - e - f - g = x$  
$a \times b \times c \times d \times e \times f \times g = x$

How should I choose the numbers?

A:

  If we choose the values so that the absolute value is necessary for the subtraction, then:
 $$a+b+c+d+e+f+g = -(a-b-c-d-e-f-g)\\2a=0$$
 Which makes $a$ zero, which is not allowed.
 So we have
 $$a+b+c+d+e+f+g = a-b-c-d-e-f-g\\b+c+d+e+f+g=0$$
 and
  $$a+b+c+d+e+f+g = a\cdot b \cdot c \cdot d \cdot e \cdot f \cdot g\\ a = a\cdot b \cdot c \cdot d \cdot e \cdot f \cdot g\\ b \cdot c \cdot d \cdot e \cdot f \cdot g=1$$
 There are many solutions to these two equations for $b,c,d,e,f,g$. We can choose $b=c$ and $d=e=f=g$ to reduce it to two equations with two variables.
 (Side remark: I tried putting $c=d=e=f=g$ first, but that leads to complex numbers, as did trying $b=c=d, e=f=g$.)
 So we now get:
 $$2b+4d=0$$
 and 
 $$b^2\cdot d^4=1$$
 Solving this gives
 $$ 4d^2 \cdot d^4 = 1\\ d=2^{-\frac{1}{3}} \text{ and } b=-2\cdot 2^{-\frac{1}{3}}$$
 So one solution is:
 $a$ has any value
 $b=c= -2\cdot 2^{-\frac{1}{3}}$
 $d=e=f=g=2^{-\frac{1}{3}}$     

A:

From the first two equations we get

 $$\begin{align}b+c+d+e+f+g &= -(b+c+d+e+f+g) \\ \Rightarrow \, b+c+d+e+f+g &= 0\end{align}$$
 As Athin already posted in his answer the second equation written with the absolute values would imply $a = 0$ and is therefore not allowed.

In order to also fulfill the third equation we need

 $b \cdot c \cdot d \cdot e \cdot f \cdot g = 1$, which is solved for example by setting $b = c = d = e = \frac{1}{A}$ and $f = g = A^2$.

Together with the condition for the first two equations we then have

 $$\begin{align}4 \cdot \frac{1}{A} + 2 \cdot A^2 &= 0 \\ \Rightarrow \, A &= -\sqrt[3]{2}\end{align}$$

Altogether, one of the many solutions is

 $a = 42 \\ b = c = d = e = -2^{-\frac{1}{3}} \\ f = g = 2^{\frac{2}{3}}$  so $x = 42$.

