For $t>1$, the distribution function of $(X+Y)/X$ is given by \begin{eqnarray*} P((X+Y)/X < t) &=&P(Y/(t-1)< X)\cr &=&\int_0^\infty P(y/(t-1)< X)\ ae^{-ay}\ dy\cr &=&\int_0^\infty e^{-ay/(t-1)} \ ae^{-ay}\ dy\cr &=&(t-1)/t. \end{eqnarray*}
Here, $S:=X+Y$ and $Z$ are independent and the density of $S$ is $g(s)=a^2 s e^{-as}$ for $s>0$ and zero otherwise (gamma density). Thus, for $t>0$, the distribution function of $(X+Y)/Z$ is given by \begin{eqnarray*} P(S/Z < t) &=&P(S/t< Z)\cr &=&\int_0^\infty P(s/t< X) \ a^2 s e^{-as}\ ds\cr &=&\int_0^\infty e^{-as/t} \ a^2se^{-as}\ ds\cr &=&[t/(t+1)]^2. \end{eqnarray*}
I want to add a comment on Michael's answer and your response. I guess that your book gave you a formula for the density of the sum of independent random variables that looks like this: $f_{X+Y}(s)=\int_{-\infty}^{\infty} f_X(s-y)\ f_Y(y)\ dy.$
You thought, "both my random variables are exponential, so I should plug in $a e^{-a(s-y)}\ ae^{-ay}$. "
But the density of an exponential random variable is not $f(y)=a e^{-ay}$, it is $f(y)=\cases{ae^{-ay} & \text{ if }y>0\cr 0 & \text{ otherwise}.}$
I often have a hard time convincing students that these two formulas are not the same. The "otherwise zero" part of the formula is crucial.
So the expression $f_X(s-y)f_Y(y)$, to be integrated over all $y$ values, is $a e^{-a(s-y)} a e^{-ay}$ only for those $y$ that satisfy $y>0$ and $s-y>0$. Otherwise, $f_X(s-y)f_Y(y)$ is zero.
In particular, the whole integral becomes zero unless $s>0$.
When $s>0$, the terms with exponent $y$ cancel each other, so the integral is quite easy $f_{X+Y}(s)=\int_0^s a^2 e^{-as} dy=a^2 s e^{-as}.$