Without loss of generality (WLOG) note that if a point in the first quadrant satisfies the equation, so do its corresponding points in the other three quadrants. Therefore, we can assume that $x, y \ge 0$, which implies that $|x|=x$ and $|y|=y$, and multiply by $4$ at the end.
We can rearrange the equation to get $x^2-x+y^2-y=0 \Rightarrow (x-\tfrac12)^2+(y-\tfrac12)^2=(\tfrac{\sqrt2}{2})^2$, which describes a circle with center $(\tfrac12, \tfrac12)$ and radius $\tfrac{\sqrt2}{2}.$ It's clear we now want to find the union of four equal areas.
[asy]draw((0,-1.5)--(0,1.5),EndArrow);draw((-1.5,0)--(1.5,0),EndArrow);draw((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,dotted); for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(1/2),-45,135));dot(rotate(i*90,(0,0))*(1/2,1/2));}[/asy]
There are several ways to find the area, but note that if you connect $(0, 1)$ to its other three respective points in the other three quadrants, you get a square of area $2$, along with four half-circles of diameter $\sqrt{2}$, for a total area of $2+2\cdot(\tfrac{\sqrt2}{2})^2\pi = \pi + 2$ which is $\boxed{\textbf{(B)}}$.

Consider the case when $x \geq 0$, $y \geq 0$.
$$x^2+y^2=x+y$$
$$(x - \frac{1}{2})^2+(y - \frac{1}{2})^2=\frac{1}{2}$$
Notice the circle intersect the axes at points $(0, 1)$ and $(1, 0)$. Find the area of this circle in the first quadrant. The area is made of a semi-circle with radius of $\frac{\sqrt{2}}{2}$ and a triangle:
$$A = \frac{\pi}{4} +\frac{1}{2}$$
[asy]draw((0,-1.5)--(0,1.5),EndArrow);draw((-1.5,0)--(1.5,0),EndArrow);draw((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,dotted);draw(arc((1/2,1/2),sqrt(2)/2,135, 315),dotted);for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(2)/2,-45,135)); dot(rotate(i*90,(0,0))*(1/2,1/2));}[/asy]
Because of symmetry, the area is the same in all four quadrants.
The answer is $\boxed{\textbf{(B)}\ \pi + 2}$