Integrate by parts:
\begin{align} \int_x^{x+1}\sin(e^t)dt & =\int_x^{x+1}e^{-t}d(-\cos(e^t)) \\ & =e^{-x}\cos e^x-e^{-x-1}\cos e^{x+1}-\int_x^{x+1}e^{-t}\cos e^{t}dt\\ & =e^{-x}\cos e^x-e^{-x-1}\cos e^{x+1}-\int_x^{x+1}e^{-2t}d\sin e^{t}\\ & =e^{-x}\cos e^x-e^{-x-1}\cos e^{x+1}-e^{-2(x+1)}\sin e^{x+1}\\ & \hphantom{={}}+e^{-2x}\sin e^x+2\int_x^{x+1}e^{-2t}\sin e^tdt.\end{align}
From here we see that $e^x \int_x^{x+1}\sin(e^t)dt$ is bounded by $1+1/e+O(e^{-x})$ and $1+1/e\approx 1.368$ can not be improved, since both $\cos e^x$ and $-\cos e^{x+1}$ may be almost equal to 1: if $e^x=2\pi n$ for large integer $n$, then $e^{x+1}=2\pi e n$, we want this to be close to $\pi+2\pi k$, i.e., we want $en$ to be close to $\frac12+k$.
This is possible since $e$ is irrational. Moreover, $e$ is so special number that you may find explicit $n$ for which $en$ is nearly half-integer: $n=m!/2$ for large even $m$ works. Indeed, $e=\sum_{i=0}^{m-1}1/i!+1/m!+o(1/m!)$ yields $em!/2=\text{integer}+1/2+\text{small}$.
