I have solved other problems like this using integration by parts. In this case, I can't figure out what to make each part for the integration. The question is true/false. Ultimately to show this you have to find the adjoint and see if it is what they claim it is, i.e., $\displaystyle uv - \int vdu$.
What do you make $u$ and $v$?
If $\langle f \vert g \rangle = \displaystyle \int_0^{\infty} f(x) g(x) e^{-x} dx$ for functions $f,g \in L_2([0,\infty))$ and $L = \left(x + \dfrac{d}{dx} \right)$ (assume that all elements of $L_2([0,\infty))$ are differentiable), its adjoint is $L^* = \left(x - \dfrac{d}{dx} \right)$.
We want $L^*$ such that $$\langle f \vert Lg \rangle = \langle L^*f \vert g \rangle$$ We have $$S=\int_0^{\infty} e^{-x}f(x)\left(x+\dfrac{d}{dx}\right)g(x) dx = \int_0^{\infty}x e^{-x}f(x) g(x) dx + \int_0^{\infty} e^{-x} f(x) dg(x)$$ We have $$\int_0^{\infty} e^{-x} f(x) dg(x) = \int_0^{\infty} d\left(e^{-x} f(x)g(x)\right) - \int_0^{\infty} g(x) d\left(e^{-x} f(x)\right)$$ Hence, $$\int_0^{\infty} g(x) d\left(e^{-x} f(x)\right) = \int_0^{\infty} g(x)\left(e^{-x} \dfrac{df}{dx} - e^{-x} f(x)\right) dx$$ Hence, $$S = \int_0^{\infty}e^{-x}\overbrace{\left(xf(x)-\dfrac{df}{dx}+f(x)\right)}^{L^*f}g(x)dx + \int_0^{\infty} d\left(e^{-x} f(x)g(x)\right)$$ Hence, the adjoint is $$L^* = \left(x+1 - \dfrac{d}{dx}\right)$$