how to solve system of linear equations of XOR operation?
As I wrote in my comment, you can just use any method you know for solving linear systems, I will use Gauss:
$$ \begin{array}{cccc|c||l} \hline x & y & z & w &\ & \\ \hline\hline 1 & 1 & 1 & 0 & 1 & \\ 1 & 1 & 0 & 1 & 1 & \text{$+$ I}\\ 1 & 0 & 1 & 1 & 0 & \text{$+$ I}\\ 0 & 1 & 1 & 1 & 1 & \\ \hline 1 & 1 & 1 & 0 & 1 & \\ 0 & 0 & 1 & 1 & 0 & \text{III}\\ 0 & 1 & 0 & 1 & 1 & \text{II}\\ 0 & 1 & 1 & 1 & 1 & \text{$+$ III}\\ \hline 1 & 1 & 1 & 0 & 1 & \\ 0 & 1 & 0 & 1 & 1 & \\ 0 & 0 & 1 & 1 & 0 & \\ 0 & 0 & 1 & 0 & 0 & \text{$+$ III}\\ \hline 1 & 1 & 1 & 0 & 1 & \\ 0 & 1 & 0 & 1 & 1 & \\ 0 & 0 & 1 & 1 & 0 & \\ 0 & 0 & 0 & 1 & 0 & \\\hline \end{array} $$ Now we can conclude $w = 0$ from line 4, which gives $z = 0$ from 3 and $y = 1$ from 2, and finally $x = 0$. So $(x,y,z,w) = (0,1,0,0)$ is the only solution.
I'll try it the 'artisanal way'.
These four rules should be enough : $\begin{align} &x\oplus 0=x\\ &x\oplus 1=\overline{x}\\ &x\oplus x=0\\ &x\oplus \overline{x}=1\\ \end{align}$
$\tag{1}1=x\oplus y\oplus z$
$\tag{2}1=x\oplus y\oplus w$
$\tag{3}0=x\oplus w\oplus z$
$\tag{4}1=w\oplus y\oplus z$
Applying $x\oplus$ on the three first equations and $w\oplus$ on the last gives :
$\tag{5}\overline{x}=y\oplus z$
$\tag{6}\overline{x}=y\oplus w$
$\tag{7}x= w\oplus z$
$\tag{8}\overline{w}=y\oplus z$
$(6)\oplus(7)\ $ gives $\ \tag{6'}1=y\oplus z$
so that from $(8)$ $w=0$
from $(7)$ and $(6)$ $\ x=z$ and $\ \overline{x}=y$
from $(5)$ $\ \overline{x}=1$ and the final conclusion :
$$(w,x,y,z)=(0,0,1,0)$$
as found (earlier) by martini...
As others have noted, all the usual methods of solving systems of linear equations (such as Gaussian elimination) in the field of real numbers work just as well in the finite field of integers modulo 2, also known as $GF(2)$.
In this field, addition corresponds to the XOR operation, while multiplication corresponds to AND (as it does in the reals, if the operands are restricted to $0$ and $1$). As both $0$ and $1$ are their own additive inverses in $GF(2)$ (since $0 \oplus 0 = 1 \oplus 1 = 0$), subtraction is also equivalent to XOR, while division is trivial (dividing by $1$ does nothing, dividing by $0$ is undefined).
So, let's try solving your example equations. Since martini already did Gaussian elimination, let me do Gauss–Jordan:
$$\begin{array}{rcrcrcrcr} 1x & \oplus & 1y & \oplus & 1z & \oplus & 0w & = & 1 \\ 1x & \oplus & 1y & \oplus & 0z & \oplus & 1w & = & 1 \\ 1x & \oplus & 0y & \oplus & 1z & \oplus & 1w & = & 0 \\ 0x & \oplus & 1y & \oplus & 1z & \oplus & 1w & = & 1 \\ \end{array}$$
The first coefficient of the first equation is already $1$, so we just subtract (XOR) that equation from the second and the third:
$$\begin{array}{rcrcrcrcr} 1x & \oplus & 1y & \oplus & 1z & \oplus & 0w & = & 1 \\ 0x & \oplus & 0y & \oplus & 1z & \oplus & 1w & = & 0 \\ 0x & \oplus & 1y & \oplus & 0z & \oplus & 1w & = & 1 \\ 0x & \oplus & 1y & \oplus & 1z & \oplus & 1w & = & 1 \\ \end{array}$$
Now the second coefficient of the second equation is $0$, so we need to choose some later row that has a $1$ for that coefficient and swap those rows:
$$\begin{array}{rcrcrcrcr} 1x & \oplus & 1y & \oplus & 1z & \oplus & 0w & = & 1 \\ 0x & \oplus & 1y & \oplus & 0z & \oplus & 1w & = & 1 \\ 0x & \oplus & 0y & \oplus & 1z & \oplus & 1w & = & 0 \\ 0x & \oplus & 1y & \oplus & 1z & \oplus & 1w & = & 1 \\ \end{array}$$
... and then subtract that row from all the others with a non-zero second coefficient:
$$\begin{array}{rcrcrcrcr} 1x & \oplus & 0y & \oplus & 1z & \oplus & 1w & = & 0 \\ 0x & \oplus & 1y & \oplus & 0z & \oplus & 1w & = & 1 \\ 0x & \oplus & 0y & \oplus & 1z & \oplus & 1w & = & 0 \\ 0x & \oplus & 0y & \oplus & 1z & \oplus & 0w & = & 0 \\ \end{array}$$
Then we subtract the third row from those with a non-zero third coefficient:
$$\begin{array}{rcrcrcrcr} 1x & \oplus & 0y & \oplus & 0z & \oplus & 0w & = & 0 \\ 0x & \oplus & 1y & \oplus & 0z & \oplus & 1w & = & 1 \\ 0x & \oplus & 0y & \oplus & 1z & \oplus & 1w & = & 0 \\ 0x & \oplus & 0y & \oplus & 0z & \oplus & 1w & = & 0 \\ \end{array}$$
... and finally the fourth row from those with a non-zero fourth coefficient:
$$\begin{array}{rcrcrcrcr} 1x & \oplus & 0y & \oplus & 0z & \oplus & 0w & = & 0 \\ 0x & \oplus & 1y & \oplus & 0z & \oplus & 0w & = & 1 \\ 0x & \oplus & 0y & \oplus & 1z & \oplus & 0w & = & 0 \\ 0x & \oplus & 0y & \oplus & 0z & \oplus & 1w & = & 0 \\ \end{array}$$
And now we can read out the result: $x = 0$, $y = 1$, $z = 0$, $w = 0$.