Assume that $\Theta$ has at least three distinct points $\theta_1$, $\theta_2,$ and $\theta_3$.
Suppose that $$\exp(\eta(\theta)T(x)+\xi(\theta)) h(x)={1\over 2}\exp(-|x-\theta|).$$
Since $h(x)$ never takes the value zero, we can write it as $h(x)={1\over 2}\exp(w(x))$,
and deduce that, for all $x\in\mathbb{R}$ and $\theta\in\Theta$,
$$\eta(\theta)T(x)+\xi(\theta) +w(x)= -|x-\theta|.$$
Substitute $\theta_1, \theta_2$ and subtract the two equations to
get
$$[\eta(\theta_1)-\eta(\theta_2)]\ T(x)+\xi(\theta_1)-
\xi(\theta_2) = |x-\theta_2|-|x-\theta_1|.$$
Since the right hand side is not a constant function of $x$, we find
that $\eta(\theta_1)\neq\eta(\theta_2)$ and hence that $T$ is differentiable
in $x$, except possibly at $\theta_1$ and $\theta_2$. The same
argument using the pairs $\{\theta_1 ,\theta_3\}$ and $\{\theta_2 ,\theta_3\}$
shows that $T$ is, in fact, differentiable everywhere.
We conclude that $|x-\theta_2|-|x-\theta_1|$ is everywhere differentiable in $x$, which is a contradiction.
There are multiple formulations of an exponential family. But whichever one chooses to follow, the basic description is that if a random variable $X\sim p_{\theta}$ where $p_{\theta}$ is a probability model (pdf or pmf), then the family of distributions $P=\{p_{\theta}:\theta\in\Omega\}$ is a one-parameter exponential family (here $\theta$ is a scalar parameter) if $p_{\theta}$ can be expressed as
$$p_{\theta}(x)=\exp\{\eta(\theta)T(x)-B(\theta)\}h(x)\quad,\,x\in\mathscr{X}\,,\tag{*}$$
where $\mathscr{X}(\subseteq \mathbb R)$ is independent of $\theta$ and $\Omega$ is some (non-degenerate) subset of $\mathbb R$.
Here $h,T$ are functions of $x$ only and $\eta,B$ are functions of $\theta$ only.
The pdf $f(x\mid\theta)$ in the question is a Gumbel density with unit scale and (unknown) location $\theta$.
We have $$f(x\mid\theta)=\exp\{\eta(\theta)T(x)-B(\theta)\}h(x)\quad,\,x\in\mathbb R\quad,\theta\in\mathbb R,$$
where $\eta(\theta)=-e^{\theta},\,B(\theta)=-\theta,\,T(x)=e^{-x}$ and $h(x)=e^{-x}$.
So this is definitely a member of a one-parameter exponential family.
In fact if the scale parameter $\sigma$ (say) is known, then the general location-scale Gumbel pdf given by $$p(x)=\frac{1}{\sigma}e^{-(x-\theta)/\sigma}\exp\left(-e^{-(x-\theta)/\sigma}\right)\qquad,\,x\in\mathbb R\quad,\theta\in\mathbb R,\sigma>0$$
also belongs to the exponential family by the same logic.
If the scale $\sigma$ is unknown, then clearly $p(\cdot)$ no longer remains in the exponential family. This is because we cannot find a $T(x)$ and an $h(x)$ in the form $(*)$ which is free of $\sigma$.
Best Answer
I just finished a similar question and I had $$f(y;p)=\exp\{y \ln(p)-\ln[\ln(1-p)]+\ln(y)\}.$$ It is the same as yours, but I put the minus sign in the denominator of the probability distribution. Note that $\theta$ is multiplying y in the exponential form, so that $\theta= \ln(p)$, and thus $p=e^\theta$. Note also that $b(\theta)=\ln[\ln(1-p)]$, and $\phi=1$ in this case. In the form that I wrote, $c(y, \phi)=\ln(y)$. I hope this helps.