$\newcommand{\la}{\lambda}$
Welcome to MathOverflow!
This conjecture is of course false, if the informal term "drawn" you are using means independence. Indeed, by homogeneity, without loss of generality $\la_2=1$. Let now $\la:=\la_1$, $(U,V,X,Y):=(a,b,c,d)$,
\begin{equation*}
S:=\frac{UX+VY}{X+Y},\quad T:=\frac{U+V}2.
\end{equation*}
We have to show that the pdfs $f_S$ and $f_T$ of $S$ and $T$ differ from each other. The random variable (r.v.) $T$ has the Gamma distribution with parameters $2,\la/2$. So,
\begin{equation*}
f_T(t)=4t\,e^{-2t/\la}/\la^21_{t>0},
\end{equation*}
whence
\begin{equation*}
f_T(t)/t\to4/\la^2\quad\text{ as }\quad t\downarrow0.
\end{equation*}
On the other hand, solving the equation $s=\frac{ux+vy}{x+y}$ for $u$, to get $u=\frac{(x+y)s-vy}x$ and $\frac{\partial u}{\partial s}=\frac{x+y}x$, we see that
the joint pdf of $(S,V,X,Y)$ is given by
\begin{align*}
f_{S,V,X,Y}(s,v,x,y)&=f_{U,V,X,Y}(\tfrac{(x+y)s-vy}x,v,x,y)\frac{x+y}x \\
&=f_U(\tfrac{(x+y)s-vy}x)f_V(v)f_X(x)f_Y(y)\frac{x+y}x \\
&=\frac1{\la^2}\,\frac{x+y}x\,
\exp\Big\{\frac{vy-(x+y)s}{\la x}-\frac v\la-x-y\Big\}
1_{x,y,v>0,\ s>\frac{vy}{x+y}}.
\end{align*}
Next, for the pdf of $S$ we have
\begin{align*}
f_S(s)&=\iint_{x,y>0} dx\,dy\,\int_0^\infty dv\,f_{S,V,X,Y}(s,v,x,y) \\
&=\frac1{\la^2}\,\iint_{x,y>0} dx\,dy\,\frac{x+y}x\,\exp\Big\{\frac{-(x+y)s}{\la x}-x-y\Big\}\int_0^{(x+y)s/y} dv\,\exp\Big\{\frac{vy}{\la x}-\frac v\la \Big\} \\
&=\frac s{\la^2}\,\iint_{x,y>0} dx\,dy\,e^{-x-y} \frac{(x+y)^2 }{x y}\,r\left(\frac{x+y}{\la y},\frac{x+y}{\la x},s\right),
\end{align*}
where
\begin{equation*}
r(a,b,s)=\frac{e^{-a s}-e^{-b s}}{(b-a)s}
=\frac1{b-a}\,\int_a^b dz\,e^{-z s}
\end{equation*}
for positive real $b\ne a$, so that $r(a,b,s)$ is decreasing in $s$; as usual, here we let $\int_a^b:=-\int_b^a$ if $b<a$; also, $r(a,b,s)\uparrow1$ as $s\downarrow0$. So, by the monotone convergence theorem,
\begin{align*}
f_S(s)/s&\to\frac1{\la^2}\,\iint_{x,y>0} dx\,dy\,e^{-x-y} \frac{(x+y)^2 }{x y} \\
&\ge\frac1{\la^2}\,\iint_{x,y>0} dx\,dy\,e^{-x-y} \frac xy=\infty
\end{align*}
as $s\downarrow0$. Comparing this with (1), we see $f_S\ne f_T$, as claimed.
