You're on the right track, but it should be $f_X(x)$ rather than $f_X(y)$ in the third-to-last line. Also, the lower bound for the integrals should be $0$ since an exponential random variable is always positive.
It might be simpler to just use the fact that
$$ F_X(y)=1-e^{-\lambda_1y}$$
if $y>0$, then compute the $y$ integral.
Let $f$ be a continuous real-valued bounded function defined on $[0,+\infty[$.
We want to compute $\mathbb{E}\big[ f(Z) \big]$. Given that $X$ and $Y$ are independent (and both follow exponential distributions), we have:
$$ \mathbb{E}\big[ f(Z) \big] = \int_{0}^{+\infty} \int_{0}^{+\infty} f\Big(\frac{x}{y+1}\Big) \lambda_1 e^{-\lambda_{1}x} \lambda_{2}e^{-\lambda_{2}y} \; dxdy. $$
Consider the change of variables $\Phi \, : \, [0,+\infty[^2 \, \rightarrow \, [0,+\infty[^2$ defined by:
$$ \forall (x,y) \in [0,+\infty[^2, \; \Phi\big( (x,y) \big) = \Big(\frac{x}{y+1}, y\Big) = (u,v) \in [0,+\infty[^2. $$
$\Phi$ is a $\mathcal{C}^{1}$ diffeomorphism from $[0,+\infty[^2$ to itself and:
$$ \Phi^{-1}\big( (u,v) \big) = \big( u(v+1), v \big) $$
$$ \mathrm{Jac}\big( \Phi^{-1}, (u,v) \big) = \begin{bmatrix} v+1 & u \\ 0 & 1 \end{bmatrix}. $$
It follows from the change of variable theorem that:
$$
\begin{align*}
\mathbb{E}\big[ f(Z) \big] & = {} \int_{0}^{+\infty} \int_{0}^{+\infty} f(u) \lambda_1\lambda_2 e^{-\lambda_{1}u(v+1)}e^{-\lambda_{2}v} (v+1) \, du dv. \\[2mm]
& = \int_{0}^{+\infty} f(u) \Bigg( \lambda_1 \lambda_2 e^{-\lambda_1 u} \int_{0}^{+\infty} (v+1) e^{-(\lambda_{1}u + \lambda_{2})v} \, dv \Bigg).
\end{align*}$$
Also, note that:
$$
\begin{align*}
\int_{0}^{+\infty} (v+1)e^{-(\lambda_{1}u + \lambda_{2})v} dv & = {} \Bigg[-\frac{e^{-av}(av + a + 1)}{a^2} \Bigg]_{v=0}^{v=+\infty} \\
& = \frac{a+1}{a^2}.
\end{align*}$$
with $a = \lambda_{1}u + \lambda_{2}$.
Therefore, the probability density function of $Z$ (with respect to the Lebesgue measure on $[0,+\infty[$) is given by:
$$ u \in [0,+\infty[ \, \mapsto \; \lambda_{1}\lambda_{2}e^{-\lambda_{1}u} \frac{\lambda_{1}u + \lambda_{2} + 1}{(\lambda_{1}u + \lambda_{2})^2}. $$
Best Answer
We know the distributions, and hence CDF and pdf of X and Y. Further we know that the random variables are independent.
$$\begin{align}X\bot Y & \iff f_{X,Y}(x,y)=f_X(x)f_Y(y), \forall (x,y)\in {\bf X\times Y} \\[1ex] X\sim{\cal Exp}(a) & \iff \Pr(X\leq x)= (1-e^{-ax})\operatorname{\bf 1}_{[0, \infty)}(x) \\ & \iff f_X(x)= a\,e^{-ax}\operatorname{\bf 1}_{[0,\infty)}(x) \\[1ex] Y\sim{\cal Exp}(b) & \iff \Pr(Y\leq y)= (1-e^{-by})\operatorname{\bf 1}_{[0, \infty)}(y) \\ & \iff f_Y(y)= b\,e^{-by}\operatorname{\bf 1}_{[0,\infty)}(y) \end{align}$$
Thus we can evaluate using a double integral, or by conditional and total probability. The later is possible because we already know what the integral of the probability distribution function will be.
However, for completeness:
$\begin{align} \Pr(X-Y> 1) & = \Pr(X > Y+1) \\ & = \iint_{{\bf X\times Y}: x>y+1} \operatorname{d}^2 F_{X,Y}(x,y) & \text{by definition} \\ & = \int_{0}^{\infty} \int_{1+y}^{\infty} \operatorname{d}F_X(x)\operatorname{d}F_Y(y) & \text{by independence} \\ & = \int_{0}^{\infty} f_Y(y) \int_{1+y}^{\infty} f_X(x) \operatorname{d}x \operatorname{d}y & \text{by expansion} \\ & =\int_{0}^{\infty} f_Y(y) \Pr(X> 1+y)\operatorname{d} y & \text{by definition} \\ & \color{gray}{= \int_{0}^{\infty} f_Y(y) \Pr(X> 1+Y\mid Y=y) \operatorname{d} y} & \color{gray}{\text{by independence; just to note}} \\ & = \int_0^\infty (b\, e^{-by})(e^{-a(1+y)}) \operatorname{d}y & \text{by substitution from the CDF and pdf} \\ & = b\,e^{-a}\, \int_0^\infty e^{-(a+b)y}\operatorname{d}y & \text{by rearranging to simplify} \\ & = b\,e^{-a}\, \left[ -\frac{ e^{-(a+b)y} }{ a+b } \right]_{y=0}^{y\to\infty} & \text{by integration} \\ & = \frac{b\,e^{-a}}{a+b} & \text{by evaluation} \\[1ex] \therefore\quad &\boxed{ \Pr(X-Y> 1)=\dfrac{b\,e^{-a}}{a+b} } \end{align}$