Is this function represented by a convolution? I was wondering whether there is an explicit $f$ such that
$\ln(\cosh(x-y)) = \int_{\mathbb{R}} |x-z|f(z-y) dz.$ 
Why do I believe that this holds?
Because the function $f(a):=\frac{1}{ \cosh^2(a)}$ already seems to come close
see wolframalpha
 A: We can change variables to $z-y$ in the integral and just assume that $y=0$ in what follows.
The answer is in fact no: putting $x=0$ gives
$$ \log{\cosh{0}} = \int_{-\infty}^{\infty} \lvert z \rvert f(z) \, dz $$
So we have
$$ \int_{-\infty}^{\infty} \lvert z \rvert f(z) \, dz = \int_{0}^{\infty} z (f(z)+f(-z)) \, dz = 0. $$
Differentiating once and evaluating at zero gives
$$ 0 = \frac{\sinh{0}}{\cosh{0}} = \int_{-\infty}^{\infty} \frac{z}{\lvert z \rvert} f(z) \, dz = \int_0^{\infty} (f(z)-f(-z)) \, dz. $$
Differentiating twice gives
$$ \frac{1}{\cosh^2{x}} = 2f(x) $$
since the derivative of the sign function is $2\delta_0$. Substituting this back in, the second equation is satisfied, but the first is not:
$$ \int_{0}^{\infty} \frac{z}{\cosh^2{z}} \, dz = \int_0^{\infty} (1-\tanh{z}) \, dz = \log{2} \neq 0. $$
So you can have
$$ \log{2}+\log{\cosh{x}} = \int_{-\infty}^{\infty} \lvert x-z \rvert \frac{1}{2\cosh^2{z}} \, dz, $$
but not without the $\log{2}$: note the convolution operator is linear, but there is no function for which $\int_{-\infty}^{\infty} \lvert x-z \rvert g(z) \, dz = C $: the left-hand side has a nonzero derivative.
