Simplify $y(t)=e^{-t}u(t) * \sum_{k=-\infty}^{\infty}\delta(t-3k)$ in the form of $y(t)=Ae^{-t}$ for $0\le t<3$

Pretty sure you just assumed the sum meant it was already written as a convolution. This is the convolution of a sum with $e^{-t}u(t)$:

$y(t)=\int_{-\infty}^{\infty} (e^{-\tau}u(\tau)\sum_{k=-\infty}^{\infty} \delta(t - \tau -3k))d\tau$

$y(t)=\int_{-\infty}^{\infty} (e^{-(t-\tau)}u(t-\tau)\sum_{k=-\infty}^{\infty} \delta(\tau -3k)) d\tau$ by commutativity

$y(t)=e^{-t} \int_{-\infty}^{\infty} (e^{\tau}u(t-\tau)\sum_{k=-\infty}^{\infty}\delta(\tau -3k)) d\tau$

$y(t)=e^{-t} \int_{-\infty}^{t} e^{\tau} \sum_{k=-\infty}^{\infty}\delta(\tau -3k)d\tau$

$y(t)=e^{-t} \int_{-\infty}^{t} \sum_{k=-\infty}^{\infty} e^{\tau} \delta(\tau -3k)d\tau$

$y(t)=e^{-t} \int_{\infty}^{t} e^0 \delta(\tau) + e^{-3} \delta(\tau + 3) + e^{-6} \delta(\tau + 6 ) + ...d\tau$ for $0 < t < 3$

$y(t)=e^{-t} (e^0 + e^{-3} + e^{-6} + ...)$

This is an infinite geometric sum:

$A = \sum_{i=0}^{\infty} (e^{-3})^{i} = \frac{1}{1-e^{-3}}$