Technically, except the very special function @Tyma Gaidash mentioned in comments, there is no closed form.
However, doing the same as @Laplacian in this post, we can write
$$\int e^{\sin(x)-x}\,dx=\sum_{n=0}^\infty \frac 1{n!} \int e^{-x} \sin ^n(x)\,dx$$
$$I_n=\int e^{-x} \sin ^n(x)\,dx$$ $$I_n=\frac{e^{-x} \left(e^{2 i x}-1\right) \sin ^n(x)}{1+i n}\,\, _2F_1\left(1,\frac{2+n+i}{2}
;\frac{2-n+i}{2} ;e^{2 i x}\right)$$ where appears the Gaussian hypergeometric function.
This could look impressive but the expressions are quite simple. They all write
$$I_n=-\frac {e^{-x}} {a_n} J_n$$ where the first $a_n$ are
$$\{1,2,10,40,680,416,20128,41600,5233280\cdots\}$$ which is not a known sequence.
The first $J_n$ are
The first integrals are
$$\left(
\begin{array}{cc}
n & J_n \\
0 & 1 \\
1 & \cos (x)+\sin (x) \\
2 & -\cos (2 x)+2 \sin (2 x)+5 \\
3 & 15 \cos (x)-3 \cos (3 x)+15 \sin (x)-\sin (3 x) \\
4 & -68 \cos (2 x)+5 \cos (4 x)+136 \sin (2 x)-20 \sin (4 x)+255 \\
5 & 130 \cos (x)-39 \cos (3 x)+5 \cos (5 x)+130 \sin (x)-13 \sin (3 x)+\sin (5 x)
\end{array}
\right)$$
Using only the terms in the table
$$\sum_{n=0}^5 \frac 1{n!} \int_0^\pi e^{-x} \sin ^n(x)\,dx=\frac{15591}{8840}-\frac{5833 e^{-\pi }}{8840}=1.73517$$ while numerical integration leads to $1.73552$. Adding the two next terms would give $1.73552$.