$$ \sum_{n=1}^{\infty}a(n)f(n/x)= \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}G((s)F(s)\frac{ds}{x^{s}} $$
where $ G(s)= \sum_{n=1}^{\infty}a(n) n^{-s}$
and $ F(s)= \int_{0}^{\infty}dx x^{s-1}f(x) $
in case $ H(t-1) =f(1/t) $ we recover Perron's formula
Do you know the distribution theory ?
$G(s)$ is the Laplace transform of the distribution $$A(t) = \sum_{n=1}^\infty a(n) \delta(t-\ln n)$$ where $\delta$ is the Dirac delta
With $B(t) = f(e^{-t}) $ : $$\sum_{n=1}^\infty a(n) f(n/e^t) = \sum_{n=1}^\infty a(n) B(t-\ln n) = A \ast B(t)$$
$F(s)$ is the Laplace transform of $B(t)$
By the convolution theorem for the Fourier/Laplace/Mellin transform, $G(s) F(s)$ is the Laplace transform of $A \ast B(t)$
By inverse Laplace transform, if everything converges $$A \ast B(t) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}G(s)F(s)e^{st}ds$$