Finding density of $U = \frac{X}{X + Y}$ for $X, \ Y $ ~ $\text{Exp}(\lambda)$ i.i.d

Hint:

Consider $U=X+Y$ and $V=\frac{X}{X+Y}$. $U$ and $V$ have joint distribution given by $$f_X(X(u,v))f_Y(Y(u,v)) J_\Phi(u,v)$$

where $J_\Phi(u,v)$ is the Jacobian determinant of the transformation $\Phi(u,v)=(uv,u-uv)$, and $f_X$, $f_Y$ are the density functions of $X$ and $Y$ (exponentials in your case)

Let $U={\large{\frac{X}{X+Y}}}$.

As you noted, $$ \frac{x}{x+y}\le u \iff y\ge \frac{x(1-u)}{u} $$ Using basically the same approach as in your attempt (and with some help from your comments), $F_U(u)$ can be computed as follows . . . \begin{align*} F_U(u) &= \int_0^\infty \int_{{\Large{\frac{x(1-u)}{u}}}}^\infty f_X(x)\,f_Y(y) \;dy \;dx \\[4pt] &= \int_0^\infty f_X(x) \left( \int_{{\Large{\frac{x(1-u)}{u}}}}^\infty f_Y(y) \;dy \right) \;dx \\[4pt] &= \int_0^\infty f_X(x) \,\left(1-F_Y\Bigl(\frac{x(1-u)}{u}\Bigr)\right) \;dx \\[4pt] &= \int_0^\infty \left( \left(\lambda e^{-\lambda x}\right) \left( e^ { -\lambda \left( {\Large{\frac{x(1-u)}{u}}} \right) } \right) \right) \;dx \\[4pt] \end{align*} which evaluates to $u$, hence \begin{align*} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\! F_U(u)\, &= \begin{cases} 0&\text{if}\;\,u\le 0\\[4pt] u&\text{if}\;\,0 < u < 1\\[4pt] 1&\text{if}\;\,u\ge 1\\[4pt] \end{cases} \\[10pt] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\! f_U(u) &= \begin{cases} 1&\text{if}\;\,0 < u < 1\\[4pt] 0&\text{otherwise}\\[4pt] \end{cases} \\[4pt] \end{align*}

We want to translate cordinates from $X,Y$ to $X,U$ where $U=X/(X+Y)$, this implies $Y=X(1/U-1)$.

So the Jacobian matrix, and its absolute determinant, are: $$\begin{align}\mathcal J(x,u)&=\dfrac{\partial\langle x, x(1/u-1)\rangle}{\partial\langle x,u\rangle}\\[1ex]&=\begin{bmatrix}\partial x/\partial x & \partial x/\partial u\\ \partial(x(1/u-1))/\partial x& \partial(x(1/u-1))/\partial u\end{bmatrix}\\[1ex]&=\begin{bmatrix}1 & 0\\ (1/u-1)& -x/u^2\end{bmatrix}\\[2ex]\lVert\mathcal J(x,u)\rVert&=\lvert x\rvert/u^2\end{align}$$

Now the support for the $X,U$ distribution is $\{\langle x,u\rangle: 0<x, 0<x(1/u-1)\}\\=\{\langle x,u\rangle: 0<x, 0<u<1\}$

Which means $x$ has a strictly positive support.

Thus the probability density function is evaluated as:

$$\begin{align}f_{\small X,U}(x,u) &= \lVert\mathcal J(x,u)\rVert f_{\small X,Y}(x, x(1/u-1))\\[1ex]&=\lambda^2~x~\mathrm e^{-\lambda x/u}/u^2\cdot \mathbf 1_{0<x, 0<u<1}\\[2ex]f_{\small U}(u) &=\dfrac{\lambda^2~\mathbf 1_{ 0<u<1}}{u^2\qquad}\int_0^\infty x\,\mathrm e^{-x\lambda/u}\mathrm d x\\[1ex]&=\mathbf 1_{0<u<1} \end{align}$$

Therefore $U$ has a standard continuous uniform distribution. $U\sim\mathcal U(0..1)$, so:

$$F_{\small U}(u)= u\,\mathbf 1_{0\leqslant u<1}+\mathbf 1_{1\leqslant u}$$