Walking a fine line

Note first that $xy = x + y$ is equivalent to $(x - 1)(y - 1) = 1$ , which is the equation of a hyperbola which divides the square $-1 \lt x \lt 1, -1 \lt y \lt 1$ into two parts.

The first part includes the corner $(-1,-1)$ , where $xy - x - y = 3 \gt 0$ , so in this part we have $xy \gt (x + y)$ . This part is then bounded by the lines $x = -1, y = -1$ and by the curve $y = \frac{x}{x - 1} = 1 + \frac{1}{x - 1}$ over the interval $-1 \lt x \lt \frac{1}{2}$ , (since when $y = -1$ we have that $-x = x - 1 \rightarrow x = \frac{1}{2}$ ).

The area of this region is then

$\int_{-1}^{\frac{1}{2}} (1 + \frac{1}{x - 1} - (-1)) dx = 2x + \ln{|x - 1|}$

evaluated from $x = -1$ to $x = \frac{1}{2}$ , giving us an area of

$1 + \ln{\frac{1}{2}} - (-2 + \ln{2}) = 3 - 2\ln{2} = 3 - \ln{4}$ .

The second part includes the other $3$ corners of the square, and in this region we have $xy \lt (x + y)$ . Since the area of the square is $4$ , the desired probability is then

$\dfrac{3 - \ln{4}}{4}$ , making $a = 3, b = 4$ and $a + b = \boxed{7}$ .