We calculate the cdf, dealing mainly with the case $a\ge 1$.
Draw the line $y=1$. The joint density function lives in the part of the first quadrant that is below the line $y=1$.
Draw the line $x+y=a$. Note that the geometry is a little different if $0\lt a\le 1$ than if $a\gt 1$, so making two diagrams is helpful.
To find the probability that $X+Y\le a$ we integrate $\lambda e^{-\lambda x}$ over the first quadrant region that is below $y=1$ and below $x+y=a$. It is convenient to integrate first with respect to $x$. We get
$$F_{X+Y}(a)=\int_{y=0}^1 \left(\int_{x=0}^{a-y}\lambda e^{-\lambda x}\,dx\right)\,dy.$$
The inner integral is $1-e^{-\lambda(a-y)}$, that is, $1-e^{-\lambda a}e^{\lambda y}$.
Now integrating from $0$ to $1$ we get
$$F_{X+Y}(a)=1-\frac{e^{-\lambda a}}{\lambda}(e^{\lambda}-1).$$
For the density function, differentiate. We get
$$f_{X+Y}(a)=e^{-\lambda a}(e^{\lambda}-1).$$
The calculation for $0\lt a\lt 1$ is similar, except that in this case
$$F_{X+Y}(a)=\int_{y=0}^a \left(\int_{x=0}^{a-y}\lambda e^{-\lambda x}\,dx\right)\,dy, $$
so $F_{X+Y}(a)=a-\frac{e^{-\lambda a}}{\lambda}(e^{\lambda a}-1)$, and now for the density we can differentiate.
Remark: If we just want the density, the second integration is in each case unnecessary if we know how to differentiate under the integral sign.
Given a uniform $U$ on the interval $[0,1]$,
$$
2U+3
$$
is uniform on $[3,5]$.
Best Answer
I assume that $a < 0$. The marginal distributions of ${\bf X}$ is:
$$f_{\bf X}(x) = \lambda e^{-\lambda x}.$$
The conditional distribution of ${\bf Y}$ knowing ${\bf X}$ is:
$$f_{\bf Y|\bf X}(y|x) = \begin{cases} \displaystyle\frac{1}{x-a} & \text{for}~y \in [a, x]\\ 0 & \text{otherwise} \end{cases}.$$
Notice that, the conditional distribution of ${\bf Y}$ knowing ${\bf X}$ is defined as:
$$f_{\bf{Y}|\bf{X}}(y|x) = \frac{f_{\bf{X},\bf{Y}}(x,y)}{f_{\bf{X}}(x)}.$$
Therefore:
\begin{align} f_{\bf{X},\bf{Y}}(x,y) & = f_{\bf{Y}|\bf{X}}(y|x)f_{\bf{X}}(x)\\ & = \begin{cases} \displaystyle\frac{\lambda e^{-\lambda x}}{x-a} & \text{for}~y \in [a, x]\\ 0 & \text{otherwise} \end{cases}& \end{align}