A one-parameter exponential family has densities

$f(x\mid\theta)=h(x)\,c(\theta)\exp\bigl\{w(\theta)\,t(x)\bigr\},$

with a support not depending on $\theta$ and a parameter space containing an interval.

(D) qualifies. The Laplace density is

$f(x\mid\theta)=\frac{1}{2\theta}\exp\Bigl\{-\frac{|x|}{\theta}\Bigr\}, \qquad x\in\mathbb{R}, \theta>0,$

which is exactly of the required form with $h(x)=1$ , $c(\theta)=1/(2\theta)$ , $w(\theta)=-1/\theta$ and $t(x)=|x|$ . The support is all of $\mathbb{R}$ , free of $\theta$ , and $\theta$ ranges over the interval $(0,\infty)$ .

The others fail:

(A) $U(0,\theta)$ has density $\theta^{-1}\mathrm{I}_{(0,\theta)}(x)$ . The support is the parameter, so the indicator cannot be absorbed into $h(x)$ — the defining disqualification.
(B) The functional form is right ( $t(x)=\log x$ , $w(\lambda)=\lambda-1$ ), but the parameter set $\{2,3,4,\dots\}$ is a discrete set of isolated points, not an interval. The family is therefore not of full rank and is not counted as a (regular) one-parameter exponential family.
(C) Expanding the $N(\theta,\theta^{2})$ exponent,

$-\frac{(x-\theta)^{2}}{2\theta^{2}} =-\frac{x^{2}}{2\theta^{2}}+\frac{x}{\theta}-\frac12 ,$

  leaves *two* distinct  $\theta$ -dependent terms, attached to
   $x^{2}$  and to  $x$ . This is a *curved* exponential family with
  a two-dimensional natural statistic  $(x,x^{2})$ , not a
  one-parameter family. (Contrast Q134, where the variance was
   $\theta$  rather than  $\theta^{2}$  and the  $x$  term lost its
   $\theta$ , collapsing the family to one parameter.)