Chọn phương án C.

Đặt $\begin{cases}
u=x\\
v'=\mathrm{e}^x
\end{cases}\Rightarrow\begin{cases}
u'=1\\
v=\mathrm{e}^x
\end{cases}$. Khi đó $$\displaystyle\int\limits_{0}^{2}x\mathrm{e}^x\mathrm{\,d}x=x\mathrm{e}^x\bigg|_0^2-\displaystyle\int\limits_{0}^{2}\mathrm{e}^x\mathrm{\,d}x=2\mathrm{e}^2-\left(\mathrm{e}^2-1\right)=\mathrm{e}^2+1.$$
Đặt $m=\displaystyle\int\limits_{0}^{2}\left(f(x)+f'(x)-\mathrm{e}^x-1\right)\mathrm{\,d}x$.
Ta có $f(x)=x\mathrm{e}^x+m$. Khi đó $$\begin{aligned}
m&=\displaystyle\int\limits_{0}^{2}\left(f(x)+f'(x)-\mathrm{e}^x-1\right)\mathrm{\,d}x\\ &=\displaystyle\int\limits_{0}^{2}\left(x\mathrm{e}^x+m+f'(x)-\mathrm{e}^x-1\right)\mathrm{\,d}x\\
&=\displaystyle\int\limits_{0}^{2}x\mathrm{e}^x\mathrm{\,d}x+f(x)\bigg|_0^2+\left(mx-\mathrm{e}^x-x\right)\bigg|_0^2\\
&=\left(\mathrm{e}^2+1\right)+\left(x\mathrm{e}^x+m\right)\bigg|_0^2+\left(2m-\mathrm{e}^2+1-2\right)\\
&=2m+2\mathrm{e}^2\\
\Leftrightarrow-m&=2\mathrm{e}^2\Leftrightarrow m=-2\mathrm{e}^2.
\end{aligned}$$
Vậy $f(x)=x\mathrm{e}^x-2\mathrm{e}^2$. Khi đó $$\begin{aligned}
\displaystyle\int\limits_{0}^{1}f(x)\mathrm{\,d}x&=\displaystyle\int\limits_{0}^{1}\left(x\mathrm{e}^x-2\mathrm{e}^2\right)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{1}x\mathrm{e}^x\mathrm{\,d}x-2\mathrm{e}^2x\bigg|_0^1\\
&=x\mathrm{e}^x\bigg|_0^1-\displaystyle\int\limits_{0}^{1}\mathrm{e}^x\mathrm{\,d}x-2\mathrm{e}^2\\
&=\mathrm{e}-\mathrm{e}^x\bigg|_0^1-2\mathrm{e}^2\\
&=\mathrm{e}-\left(\mathrm{e}-1\right)-2\mathrm{e}^2=1-2\mathrm{e}^2.
\end{aligned}$$