Let $G \colon [0,1] \to [0,1]$ be a differentiable cumulative distribution function (monotonically non-decreasing function with $G(0) = 0$ and $G(1) = 1$). We say that $G$ is regular if $$ x - \frac{1- G(x)}{G'(x)}$$ is monotonically non-decreasing.
Let $F_1,F_2 \colon [0,1] \to [0,1]$ be regular twice differentiable cumulative distribution functions.
Must $F = F_1F_2$ be regular?
The answer is no. In the definition of the regularity, we assume that $G'>0$ on $(0,1)$ and the monotonicity in question is over $(0,1)$. Suppose for a moment that $G$ is any cumulative distribution function (cdf) on $\mathbb{R}$ such that $G=0$ on $(-\infty,0)$, $G=1$ on $[1,\infty)$, $G$ is twice differentiable on $(0,1)$, and $ x - \dfrac{1- G(x)}{G'(x)}$ is nondecreasing in $x\in(0,1)$. Let $$g:=\frac1{1-G},$$ so that $G=1-\dfrac1g$, assuming that $\frac10=\infty$ and $\frac1\infty=0$. The above conditions on $G$ can be translated into the following: $$(*)\qquad\text{$g$ is nondecreasing on $\mathbb{R}$, $g=1$ on $(-\infty,0)$, $g=\infty$ on $[1,\infty)$, }$$ $$(**)\qquad\text{and on $(0,1)$ the function $g$ is twice differentiable, with $g'>0$ and $g''\ge0$.} $$
Let $g_1(x):=2 + 5 (x - a) + \frac14 (x - a)^2$ and $g_2(x):=\frac54 + \frac12 (x - a) + \frac14 (x - a)^2$, where $a:=\frac34$. Let then $G_j:=1-\dfrac1{g_j}$ for $j=1,2$, $$(!)\qquad G:=G_1G_2,$$ and $g:=\dfrac1{1-G}$. Then $g''(a+\delta)<-0.037<0$ for $\delta:=\frac1{200}$ and hence $ x - \dfrac{1- G(x)}{G'(x)}$ is decreasing in $x$ in a neighborhood of the point $a+\delta[\in(0,1)]$.
However, $G_1$ and $G_2$ are not cdf's, and so, we need to work a bit more. Again for $j=1,2$, let $h_j$ be the function on $(-\infty,1)$ such that $$ h_j(x)= \begin{cases} g_j(x)&\text{ if }a\le x<1,\\ 1\vee[g_j(a)+g_j'(a)(x-a)]&\text{ if }x\le a. \end{cases} $$ Note that $h_j$ is nondecreasing and convex, and $h_j=1$ on $(-\infty,\frac14]$; that is, conditions $(*)$--$(**)$ are "almost" satisfied with $h_j$ in place of $g$. Also, $h_j=g_j$ on $[a,1)$. It is easy (but a bit tedious) to construct a "smooth" enough approximation $f_j\colon\mathbb{R}\to\mathbb{R}$ of $h_j$ such that $f_j=h_j$ in a neighborhood of the point $a+\delta$, conditions $(*)$--$(**)$ hold for $f_j$ in place of $g$, and $F_j:=1-\dfrac1{f_j}$ is a cdf twice differentiable on $\mathbb{R}$, with $F_j=0$ on $(-\infty,0)$ and $F_j=1$ on $[1,\infty)$. Then in a neighborhood of $a+\delta$ one has $f_j=g_j$ and hence $F:=F_1F_2=G$, where $G$ is as in $(!)$. So, $ x - \dfrac{1- F(x)}{F'(x)}$ is decreasing in $x$ in a neighborhood of the point $a+\delta[\in(0,1)]$.
Addendum: One way to to get the mentioned "smooth" approximations $f_j$ of $h_j$ is to use splines. Thus, one e.g. may choose $$ f_1(x):=1 + 5\Big(\frac15\Big)p_1\Big(\frac{x-11/20}{1/5}\Big) + \frac14\,\Big(\frac1{200}\Big)^2 p_2\Big(\frac{x-3/4}{1/200}\Big)+ \frac1{10^6}\Big(\frac{x_+}{(1 - x)_+}\Big)^3 $$ and $$ f_2(x):=1 + \frac12\,\Big(\frac15\Big)p_1\Big(\frac{x-1/4}{1/5}\Big) + \frac14\,\Big(\frac1{200}\Big)^2 p_2\Big(\frac{x-3/4}{1/200}\Big)+ \frac1{10^6}\Big(\frac{x_+}{(1 - x)_+}\Big)^3 $$ for all real $x$, where $x_+:=0\vee x$ and $$ p_1(x):=x_+^3+\frac{(x+1)_+^3}2-\frac{(x+1)_+^4}4+\frac{(x-1)_+^3}2+\frac{(x-1)_+^4}4= \begin{cases} 0 & \text{ if } x\leq -1, \\ \frac14 (1+2x+2|x|^3-x^4) & \text{ if } -1<x\leq 1, \\ x & \text{ if } x>1 \end{cases} $$ and $$ p_2(x):=3x_+^3 - 3x_+^4 + x_+^5 - (x - 1)_+^3 - 2(x - 1)_+^4 - (x - 1)_+^5= \begin{cases} 0 & \text{ if } x\leq 0, \\ x^5-3 x^4+3 x^3 & \text{ if } 0<x\leq 1, \\ x^2 & \text{ if } x>1 \end{cases} $$ are twice continuously differentiable splines, with $p_j'\ge0$ and $p_j''\ge0$ for $j=1,2$, which approximate $x_+$ and $x_+^2$, respectively.
Then, as desired, for each $j=1,2$ the function
$F_j:=1-\dfrac1{f_j}$ is a cdf, which is regular in the sense defined in the question and twice continuously differentiable on $\mathbb{R}$, with $F_j=0$ on $(-\infty,0)$ and $F_j=1$ on $[1,\infty)$.
On the other hand, for $F:=F_1F_2$, the derivative of $x - \dfrac{1- F(x)}{F'(x)}$ in $x$ at $x=a+\delta=\frac34+\frac1{200}$ is $\approx-0.016<0$, and so, $x - \dfrac{1- F(x)}{F'(x)}$ is decreasing near $x=a+\delta$.
