This is probably not unique. Due to translational invariance, we can restrict the space on which we're acting to functions $f$ on the interval $[-1/2,1/2]$, and if we specify those to be odd, $-\frac{1}{2} \frac{d^2 }{dx^2} $ seems a viable inverse:
$$
-\frac{1}{2} \frac{d^2 }{dx^2} K(x,y) = \delta (x-y)
$$
If we act with $K$ on an $f$ from our space, we again obtain a result from our space,
$$ 
\int_{-1/2}^{1/2} dy K(x,y) f(y) =
\int_{-1/2}^{1/2} dy \left[ 1-\left| x - y \right| \right] f(y) = \int_{-1/2}^{x} dy \left[ x-y \right] f(y) +\int_{x}^{1/2} dy \left[ y-x \right] f(y) =\int_{-x}^{1/2} dy \left[ x+y \right] f(-y) +\int_{-1/2}^{-x} dy \left[ -y-x \right] f(-y) = -\int_{-1/2}^{1/2} dy [1-|-x-y|] f(-y) = -\int_{-1/2}^{1/2} dy K(-x,y) f(y)
$$
Of course, the same is true for acting with the inverse. HOLD ON THIS ISN'T QUITE RIGHT YET