This follows from Weyl's equidistribution theorem. When $b$ is irrational, convergence holds whenever $f$ is continuous, or even just Riemann-integrable (see http://individual.utoronto.ca/hannigandaley/equidistribution.pdf for an exposition). Integrability is not enough by itself, since you can make the sum zero by changing $f$ on the countable set $\{a+b\ell\}$, which has measure zero.
When $b$ is rational, convergence certainly doesn't hold for most $f$, since $a+b\ell$ takes on just finitely many values modulo $1$.