Let $u=({\log }_{e}x{)}^{3/2}$, so $du=\frac{3}{2}\frac{({\log }_{e}x{)}^{1/2}}{x}dx$. The integrand transforms via $\frac{({\log }_{e}x{)}^{1/2}}{x}dx=\frac{2}{3}du$.

With $t=a-u$, $dt=-du$. When $x=1$, $u=0$, so $t=a$; when $x=e$, $u=1$, so $t=a-1$.

$I={\int }_a^{a-1}\frac{1}{t^2}\cdot \left(-\frac{2}{3}\right)dt=\frac{2}{3}{\left[\frac{1}{t}\right]}_a^{a-1}=\frac{2}{3}\left(\frac{1}{a-1}-\frac{1}{a}\right)=\frac{2}{3}\cdot \frac{1}{a(a-1)}.$

Setting $I=1$: $\frac{2}{3a(a-1)}=1\Rightarrow 3a^2-3a-2=0$.

$a=\frac{3\pm \sqrt{9+24}}{6}=\frac{3\pm \sqrt{33}}{6}.$

Both roots are irrational; numerically $a\approx 1.46$ (which lies in $(1,\infty )$) and $a\approx -0.46$ (which lies in $(-\infty ,0)$), so both satisfy the domain constraint.

$\therefore$ Options (C) and (D) are correct.