Suppose that
$$Z=\dfrac{X}{X+Y}$$
$$X \sim Gamma(a,\lambda)$$ $$ Y \sim Gamma(b,\lambda)$$
with $X$ and $Y$ independent. I would like to see if it might be possible to determine the distribution of $Z$ without resorting to Jacobians. Is it possible to do a proof by representation or by way of moment generating functions? For example, I tried using iterated expectations over
$$ \mathbb{E}\left(e^{tZ}\right) = \mathbb{E}\left(e^{t\cdot\frac{X}{X+Y}}\right) $$
where
\begin{align*} \mathbb{E}\left(\mathbb{E}\left(e^{t\cdot\frac{X}{X+Y}}\right)\bigl\vert X+Y=c\right) &= \mathbb{E}\left(\mathbb{E}\left(e^{t\cdot\frac{c-Y}{c}}\right)\right) \\ &=\mathbb{E}\left(\mathbb{E}\left(e^{t\cdot \left(1-\frac{Y}{c}\right)}\right)\right) \\ &= e^t \cdot \mathbb{E}\left(\mathbb{E}\left(e^{t \left(-\frac{Y}{c}\right)}\right)\right) \\ \end{align*}
which seems to be saying I need to find the MGF of $-\frac{Y}{c}$. Since $Y \sim Gamma(b, \lambda)$, we have that $\dfrac{Y}{c} \sim Gamma\left(b, \frac{\lambda}{c}\right)$. With the negative on $\dfrac{Y}{c}$, I am unsure how this would work. Should I instead condition on $X$ instead of $X+Y$? Or do MFGS just not work well?
The joint density is $$ \text{constant} \times x^{a-1} e^{-\lambda x} \cdot y^{\beta-1} e^{-\lambda y} $$
Now put $c-y$ in place of $x$: \begin{align} & \text{constant}\times (c-y)^{a-1} y^{b-1} e^{-\lambda(c-y)} e^{-\lambda y} \\[8pt] = {} & \text{constant} \times \left( 1 - \frac y c \right)^{a-1} \left( \frac y c \right)^{b-1} \\[8pt] = {} & \text{constant} \times (1-u)^{a-1} u^{b-1} \end{align} (The "constants" are not all equal to each other.)