Signal with Complex Gaussian noise
No, it will be Rician Distribution.

Rayleigh distribution is a special case of Rician Distribution when the normal random variables involved are of zero mean and equal variances.

I have assumed you meant :$$y=x_r + j \cdot x_i + n_r + j \cdot n_i$$ where $x_r, x_i$ are deterministic variables.

Basically, you now have $y_r = (n_r + x_r),\ \sim \mathcal N(x_r, \sigma^2) $ and $y_i = (n_i + x_i), \ \sim \mathcal N(x_i, \sigma^2)$

Now, you can write $y = y_r +j \cdot y_i = r \cdot e^{j\phi}$, where $r,\phi$ are random variables and, $y_r = r \cdot \cos(\phi)$ and $y_i = r \cdot \sin(\phi)$.

Using method of transformations, Jacobian matrix can be given as follows:
$$\begin{pmatrix} y_r \\ y_i \end{pmatrix} = \begin{pmatrix} r \cdot \cos(\phi) \\ r \cdot \sin(\phi)\end{pmatrix}$$
$$\mathbf J = \begin{pmatrix} \cos(\phi)&amp;&amp;-r \cdot \sin(\phi) \\ \sin(\phi)&amp;&amp; r \cdot \cos(\phi)\end{pmatrix}$$
$$|\mathbf J| = r$$
Hence, you see that Jacoboan does not change by changing the mean of the normal random variables from 0 to non-zero.

Joint PDF of derived RVs $r, \phi$ will be given by the following :
$$f_{R,\Phi}(r, \phi) = f_{Y_r, Y_i}(r\cos(\phi), r\sin(\phi)) \cdot |\mathbf J |$$
$$f_{R,\Phi}(r, \phi) = \frac{r}{2\pi \sigma^2} \cdot \exp\left[-\frac{(r\cos(\phi) - x_r)^2+(r\sin(\phi)-x_i)^2)}{2\sigma^2}\right]$$ 
$$\forall r\in [0,\infty) \\ \phi \in[-\pi, \pi]$$

Solve for Marginal PDF of $R$ and $\Phi$ to see that it is not Rayleigh anymore. But a more generalized Rician Distribution.