If $(X,Y)$ is binormal, then so is $(X,Z) = (X,X+Y)$. The ratio $X/Z$ is the tangent of the slope of the line through the origin and the point $(Z,X)$. When $X$ and $Z$ are uncorrelated with zero means, it is well known (and easy to compute) that $X/Z$ has a Cauchy distribution. Cauchy distributions have no expectations. This should lead us to suspect $X/Z$ might not have a mean, either. Let's see whether it does nor not.
For any angle $0 \lt \theta \lt \pi/2$, consider the event
$$E_\theta = \{(Z,X)\,|\, X \ge Z\cot(\theta\}.$$
This is of interest because its probability is the chance that $X/Z$ exceeds $\cot(\theta)$: the survival function of $X/Z$. It carries all the information of the distribution function of $X/Z$.
$E_\theta$ is a (closed) cone in the plane consisting of all points on all lines making an angle of $\theta$ or less to the right of the vertical ($X$) axis. Let's underestimate the probability of $E_\theta$. To do so, we will work in polar coordinates. Consider any possible radius $\rho$. Among all points of this radius within the set $E_\theta$, the density $f$ of $(Z,X)$ will achieve a minimum value $f_\theta(\rho)$. This minimum must be nonzero provided the density does not degenerate. (More about this possibility later.) Use this to bound the probability
$$\eqalign{
\Pr(E_\theta) &= \int_{\pi/2-\theta}^{\pi/2}\int_0^\infty f(\phi,\rho) \rho d\rho d\phi \\
&\ge \int_{\pi/2-\theta}^{\pi/2}\int_0^\infty \rho f_\theta(\rho) d\rho d\phi \\
&=\theta \int_0^\infty \rho f_\theta(\rho) d\rho \\
&= C(\theta) \theta
}$$
where I have written $C(\theta)$ for the integral, which is some positive number depending on $\theta$. Moreover, for $0\lt\theta\lt\pi/2$, $C(\theta)$ has a nonzero lower bound $C \gt 0$.
By definition, the expectation of $X/Z$ is the sum of two parts: one integral for the positive part when $X/Z \ge 0$ and another for the negative part when $X/Z \lt 0$. Let's tackle the positive part. For any positive random variable $W$ with distribution function $F$, integration by parts shows its expectation equals the integral of its survival function $1-F$, since
$$\mathbb{E}(W) = \int_0^\infty w dF(w) = (w(1-F(w))|_0^\infty + \int_0^\infty (1-F(w)) dw = \int_0^\infty (1-F(w)) dw.$$
Applying this to $W = X/Z$ and substituting $w=\cot(\phi)$ gives for the positive part of the integral
$$\eqalign{
\int_0^\infty (1 - F(w)) dw &= \int_0^{\pi/2} (1 - F(\cot(\phi))) \csc^2(\phi) d\phi \\
&= \int_0^{\pi/2} \Pr(E_\phi) \csc^2(\phi) d\phi \\
&\ge C \int_0^\theta \phi \csc^2(\phi) d\phi \\
&\gt C \int_0^\theta \frac{d\phi}{\phi}.
}$$
(The final inequality is a simple consequence of the well-known inequalities $0 \lt \sin(\phi) \lt \phi$ for $0 \lt \phi \lt \pi$, which upon taking the $-2$ power gives $\csc^2(\phi) \gt 1/\phi^2$.)
For any $\theta \gt 0$, the last term is a divergent integral, because for $0\lt \epsilon$,
$$\int_0^\theta \frac{d\phi}{\phi} \gt \int_\epsilon^\theta \frac{d\phi}{\phi} = \log(\theta) - \log(\epsilon) \to \infty$$
as $\epsilon \to 0^{+}$.
Consequently, the positive part of the expectation does not exist. It is immediate that the expectation of $X/W$ does not exist, either.
We left behind one exception to consider: when $X/Z$ is supported on a line passing through the origin, this argument breaks down (because then the density can equal zero--and in fact is zero for almost all $\theta$). In this degenerate case, $X/Z$ reduces to a constant--equal to tangent of the slope of that line--and obviously that constant is its expectation. This is the only such situation in which $X/Z$ has an expectation.
