Given $$u(t) = \int_0^t y(\tau) d\tau$$
Consider a convolution type of integral
$$W = \int_0^t\lambda^{t-\tau}y(\tau) d\tau$$
$\lambda$ a positive real number
Is it possible to write $W = f(u(t))$, where $f$ is some/any function?
i.e. is it possible to find $f$?
Well, I don't see how you can get $W(t)$ as a composed function of $u(t)$ as $W(t) = f\left(u(t)\right)$, but you can do the following, which may be of some use:
Assuming $y(t)$ is a sufficiently-smooth function, note that $$ u(0) = 0 $$ and $$ u'(t) = y(t)\, . $$ Plug the latter expression for $y(t)$ into your integral for $W(t)$: $$ W(t) = \int_0^{\tau} d\tau\, \lambda^{t-\tau} u'(\tau) $$ Integrate this by parts once to yield: \begin{align} W(t) &= {\left.\lambda^{t-\tau} u(\tau)\right|}_{\tau = 0}^{\tau = t} + \ln\lambda\int_0^t d\tau\, \lambda^{t-\tau} u(\tau) \\ &= u(t) + \ln\lambda\int_0^t d\tau\, \lambda^{t-\tau} u(\tau)\, . \end{align} This gives us $W(t)$ as a functional of $u(t)$, but not as a function. I doubt such a function exists for general $y(t)$.