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 (or better to say there is no reason why the can't).