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 $[-1/2,1/2]$. If we specify those to be odd and with vanishing derivative at the boundaries, i.e., $f(-x)=-f(x)$ and $f^{\prime } (-1/2)=f^{\prime } (1/2)=0$, then $-\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,
$$
\int_{-1/2}^{1/2} dy K(x,y) f(y) =
\int_{-1/2}^{1/2} dy \left[ 1-\left| x - y \right| \right] f(y) =
$$

$$
\int_{-1/2}^{x} dy \left[ y-x \right] f(y) +\int_{x}^{1/2} dy \left[ x-y \right] f(y) =
$$

$$
\int_{-x}^{1/2} dy \left[ -x-y \right] f(-y) +\int_{-1/2}^{-x} dy \left[ x+y \right] f(-y) =
$$

$$
\int_{-1/2}^{1/2} dy [1-|-x-y|] f(-y) = -\int_{-1/2}^{1/2} dy K(-x,y) f(y)
$$
as well as
$$
\left. \frac{d}{dx} \int_{-1/2}^{1/2} dy [1-|x-y|] f(y) \right|_{x=1/2} =
$$

$$
\left. \int_{-1/2}^{1/2} dy [ 1-2\theta (x-y)] f(y) \right|_{x=1/2} =
$$

$$
-\int_{-1/2}^{1/2} dy f(y) =0
$$
Of course, the same is true for acting with the inverse.