The solution was obtained already by <A HREF="https://www.researchgate.net/publication/303008084_Distribution_of_eigenvalues_for_some_sets_of_random_matrices">Marchkenko and Pastur,</A> in terms of the Cauchy transform $g(z)$ of the spectral density of $D+XX^{\rm T}$, see for example equations 2 + 3 of <A HREF="https://hal.archives-ouvertes.fr/hal-01346371/document">Spectrum of deformed random matrices and free probability:</A>

$$g(z)=\int\frac{1}{z-t-(m/n)(z-t)g(z)-1+m/n}\rho(t)dt$$

where $\rho(t)$ is the spectral density of $D$. This holds for any random Hermitian perturbation $D$, irrespective of whether it is diagonal or not.

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Let me try an alternative approach; The Cauchy transform of the <A HREF="https://en.wikipedia.org/wiki/Free_Poisson_distribution">Marchenko-Pastur distribution</A>, supported on $[(1-\sqrt{\lambda})^2,(1+\sqrt{\lambda})^2]$, is given by 
$$g_{\rm MP}(z)=\frac{-\sqrt{(-\lambda+z-1)^2-4 \lambda}-\lambda+z+1}{2 z}.$$
The perturbation $D$, with $\rho(t)=w\delta(t-1)+(1-w)\delta(t)$, has Cauchy transform
$$g_D(z)=\frac{w}{z-1}+\frac{1-w}{z}.$$
The Cauchy transform $g_{D+{\rm MP}}(z)$ of $D+XX^{\rm T}$ is given by the composition of free probability,
$$R_{D+{\rm MP}}=R_{\rm MP}+R_D,\;\;R(z)=z-1/g(z).$$
This gives the Cauchy transform
$$g_{D+{\rm MP}}(z)=\frac{1}{z}\left[ -\frac{2}{\sqrt{(\lambda-z+1)^2-4 \lambda}+\lambda-z-1}+\frac{z-1}{w+z-1}-1\right]^{-1}.$$

Now the goal is to integrate
$\langle\log |t|\rangle_t=\int \rho_{D+{\rm MP}}(t)\log t\,dt$, which might be obtainable directly from the Cauchy transform via
$$\frac{d}{dz}\langle\log |z-t|\rangle_t = \int \frac{\rho_{D+{\rm MP}}(t)}{z-t}\,dt=g_{D+{\rm MP}}(z).$$