I was imagining that if a system's vector is an eigenvector, with eigenvalue k, of some observable O, then any measurement of O will come out to k and will leave the system in that same eigenvector state - even if O is some kind of "compound observable".
Try it.
Define
$$\psi_0 = \exp \left( \frac{x^2}{2i}−cx \right)$$
and act on it with $x + p$ to get
$$ \begin{array} \ \psi_1 & = (x+p)\psi_0 \ & = x\exp\left(\frac{x^2}{2i}−cx\right) - i\frac{d}{dx}\exp\left(\frac{x^2}{2i}−cx\right) \ & = x\exp\left(\frac{x^2}{2i}−cx\right) - i\exp\left(\frac{x^2}{2i}−cx\right) \frac{d}{dx}\left(\frac{x^2}{2i}−cx\right) \ & = x\exp\left(\frac{x^2}{2i}−cx\right) - i\exp\left(\frac{x^2}{2i}−cx\right) \left(\frac{2x}{2i} - c\right) \ & = c\exp\left(\frac{x^2}{2i}−cx\right) = c \psi_0 \end{array}$$ which shows that $\psi_0$ is an eigenfunction of $(x+p)$ (which you already knew, but I exhibit it to show the difference with the position operator below).
So we try the same thing with just $x$, but
$$\psi_2 = x\psi_0 = x \exp\left(\frac{x^2}{2i}−cx\right) $$
is not a constant times $\psi_0$ so your your wave function is not an eigenfunction of the position operator. That result turns out to be general, $\hat{O} = \hat{x} + \hat{p}$ does not share eigenfunctions with either the position or the momentum operator.
I get the feeling that you are groping toward a method to understand these issues without the (potential) complexity of working the problem long hand every time. That's a good idea; the usual framework is to define the "commutator" between to operators $\hat{A}$ and $\hat{B}$ as
$$ \left[\hat{A},\hat{B}\right] = \hat{A}\hat{B} - \hat{B}\hat{A} \,.$$
With a little effort you can show that when the commutator is zero the operators share eigenstates, and when it is non-zero they do not.
With this in mind we could have written
$$\begin{array} \ \left[ x,(x+p) \right] & = x\left(x-i\frac{d}{dx}\right) - \left(x-i\frac{d}{dx}\right)x \ & = \left( x^2 - ix\frac{d}{dx} \right) - \left(x^2 - i\frac{d}{dx}x \right) \ & = \left( x^2 - ix\frac{d}{dx} \right) - \left(x^2 - i \right) \ & = i \left( \frac{d}{dx}x - x\frac{d}{dx} \right) \ & \ne 0 \end{array} $$
and drawn the same conclusions. 66