Impulse response of a transfer function
You're almost there; you just need to connect a few dots.  Let your $\frac{10}{5 + i 2 \pi f} = G(f)$.  Then $g(t) = 10 u(t) e^{-5 t}$.

Now we get into that exponent part.  Your $h(t) = g(t - t_0)$ is correct, but you're applying it incorrectly.  You need to apply it to the part that I've labeled as $g(t)$: $h(t) = g(t - t_0) = 10 u(t - t_0) e^{-5 (t - t_0)} = 10 u(t - 4) e^{-5 (t - 4)}$.

So, basically, it's the output you'd expect from the "pure" transfer function, just delayed by 4 time units (presumably seconds). 

Does that help?