Indeed $\overline X-\theta$ is a valid pivot for $\theta$ when $\beta$ is known. All you have to do now is find $a,b$ such that $a<\overline X-\theta<b\iff \theta\in(\overline X-b,\overline X-a)$ with the desired confidence coefficient $P_{\theta}\left[\theta\in(\overline X-b,\overline X-a)\right]$ (for a two-sided interval). This is easily done using software.
Note that you have $X_i-\theta\stackrel{\text{i.i.d}}\sim\mathsf{Exp}$ with mean $1/\beta$, which implies $X_{(1)}-\theta\sim \mathsf{Exp}$ with mean $1/(n\beta)$ where $X_{(1)}=\min\limits_{1\le j\le n}X_j$. So yet another pivotal quantity is $$T(\mathbf X,\theta)=2n\beta(X_{(1)}-\theta)\sim \chi^2_2$$
We expect a confidence interval based on this pivot to be 'better' (in the sense of shorter length, at least for large $n$) than the one based on $\sum\limits_{i=1}^n X_i$ as $X_{(1)}$ is a sufficient statistic for $\theta$.
Now you can derive a two-sided confidence interval with confidence level $1-\alpha$ starting from
$$P_{\theta}(\chi^2_{1-\alpha/2,2}< T< \chi^2_{\alpha/2,2})=1-\alpha\quad\forall\,\theta$$
Here $\chi^2_{\alpha,2}$ is of course the $(1-\alpha)$th fractile of $\chi^2_2$, i.e. $P(\chi^2_2>\chi^2_{\alpha,2})=\alpha$. Notably if you have an observed sample at hand, then calculations involving chi-square fractiles for the confidence interval can be done by hand since printed chi-square tables are readily available.