Since $X_1$, $X_2$, $X_3$, $X_4 \sim \mathcal{E}(\lambda)$ are i.i.d., the joint distribution probability distribution function (PDF) of the vector $X = (X_1, X_2, X_3, X_4)$ writes:
$$
f_X(x) = \lambda^4 \exp\left(-\lambda (x_1 + x_2 + x_3 + x_4)\right)\,\boldsymbol{1}_{x_1\geq 0,x_2\geq 0,x_3\geq 0,x_4\geq 0} \, .
$$
To show that the components $Y_1$, $Y_2$, $Y_3$, $Y_4$ of the vector $Y=g(X)$ are independent, one can show that the joint PDF $f_Y$ of $Y$ is separable. To do so, let us compute $f_Y$ by the change of variable method. We consider a bounded measurable function $h$, and compute $\mathbb{E}h(Y)$:
\begin{aligned}
\mathbb{E}(h\circ g)(X) &= \int_{\mathbb{R}^4} h\circ g(x)\, f_{X}(x) \,\mathrm{d}x \, ,\\
&= \int_{\mathbb{R}^4} h(y)\, \underbrace{f_{X}\big({g}^{-1}(y)\big) \big|J_{{g}^{-1}}(y)\big|}_{f_Y(y)} \,\mathrm{d}y \, ,
\end{aligned}
where $J_{g^{-1}}$ is the Jacobian of the reciprocal function of $g$, defined by $$ g^{-1}(y) = \big(y_1 y_2 y_3 y_4, y_1 y_3 y_4 (1 - y_2), y_1 y_4 (1 - y_3), y_1 (1-y_4)\big) . $$
We get
$$
f_Y(y) = \lambda^4\, {y_1}^3 \exp\left(-\lambda y_1 \right)\, \boldsymbol{1}_{0\leq y_1}\, \boldsymbol{1}_{0\leq y_2\leq 1}\, y_3 \, \boldsymbol{1}_{0\leq y_3\leq 1}\, {y_4}^2\, \boldsymbol{1}_{0\leq y_4\leq 1} \, ,
$$
which is in separated form. Note that this exercise is similar.