I'd like to clear up some confusion in your question. First you state that for $y_c(t)$ to be periodic, its Fourier transform should consist only of Dirac delta impulses. That's not true, at least it's not sufficient. As an example, take a signal with Fourier transform
$$Y(j\omega)=\delta(\omega -1)+\delta(\omega - \sqrt{2})\tag{1}$$
The corresponding signal is not periodic even though its spectrum only consists of Dirac delta impulses. The reason is that the discrete frequency contributions of a periodic signal must be at integer multiples of the fundamental frequency, which is not the case for the spectrum given in $(2)$. So it's important to realize that a discrete spectrum is only necessary but not sufficient for periodicity.
The second confusion is that you don't see how $Y_c(j\omega)$ can only consist of Dirac impulses. Note that for any function $H(j\omega)$ (which is continuous at $\omega=\omega_0$), we have
$$H(j\omega)\delta(\omega-\omega_0)=H(j\omega_0)\delta(\omega-\omega_0)\tag{2}$$
So no matter which form $H_0(j\omega)$ takes, $Y_c(j\omega)$ will always be an infinite sum of Dirac impulses, where the weight of each impulse is determined by the value of $H_0(j\omega)$ at the respective frequency. So the question is not whether $Y_c(j\omega)$ is a sum of weighted Dirac impulses (it always is!), but whether this sum of Dirac impulses corresponds to a periodic signal.
So it is only about the relation of the frequencies of the Dirac impulses, not about their weights, which are determined by $H_0(j\omega)$. So for the determination of periodicity, $H_0(j\omega)$ is irrelevant, as already pointed out in Marcus Müller's answer.
Since you already know that the ratio $\omega_s/\omega_0$ must be rational for the sampled signal to be periodic, let me show you how to obtain the ratio of the amplitudes of the first two non-zero Fourier coefficients of $y_c(t)$. Since $\omega_s>2\omega_0$, the first frequency component of $y_c(t)$ is at $\omega_0$, and the second one is at $\omega_s-\omega_0$. As mentioned before, the weights of the Dirac impulses are given by $H_0(j\omega)$ evaluated at the respective frequencies. So the amplitude ratio of the first and second Fourier coefficient is
$$\frac{H(j\omega_0)}{H(j(\omega_s-\omega_0))}=\frac{\text{sinc}(\frac{\omega_0}{\omega_s})}{\text{sinc}(\frac{\omega_s-\omega_0}{\omega_s})}=\frac{\sin(\pi\frac{\omega_0}{\omega_s})}{\pi\frac{\omega_0}{\omega_s}}\frac{\pi(1-\frac{\omega_0}{\omega_s})}{\sin(\pi(1-\frac{\omega_0}{\omega_s}))}=\frac{\omega_s}{\omega_0}-1\tag{3}$$
because $\sin(\pi-x)=\sin(x)$.