Suppose there are three random variables $X,Y,Z$ such that $$D_{KL}(X,Y)\leqslant D_{KL}(X,Z)$$ And there is an independent (with respect to $X,Y,Z$) Gaussian noise $\xi\sim N(0,1)$. Can we say the following?
$$ D_{KL}(X+\xi,Y+\xi)\leqslant D_{KL}(X+\xi,Z+\xi) $$
This is not true.
The first reason to expect that this doesn't hold is that one can get a sort of free pass due to the infinitude of KL divergence between discrete and continuous laws. So if we make the law of $Z$ discrete, but the law of $X$ to be a slightly smoothened version of this, and the law of $Y$ very different from that of $X$, but smooth, then after smoothing with the Gaussian the order of KL divergences could well be flipped. We can execute this idea somewhat more simplistically.
Recall that the fact that $D (\mathcal{N}(0,\sigma_1^2)\|\mathcal{N}(0,\sigma_2^2)) = \frac12\left( (\sigma_1/\sigma_2)^2 - 1- \log (\sigma_1/{\sigma_2})^2\right)$ in nats (i.e., where the log in the definition of $D$ is natural). I'll use $\gamma$ to denote the standard Gaussian law.
Let $s > 1$ be a parameter we will fix later. Take $P, Q$ to be centred Gaussians of variance $1$ and $s - 1$, and $R$ to be the point mass at $0$.
We have $D(P\|Q) < \infty = D(P\|R)$. But $P*\gamma = \mathcal{N}(0,2), Q*\gamma= \mathcal{N}(0,s),$ and $R * \gamma = \mathcal{N}(0,1)$. In this case, we get $$ D(P*\gamma\|Q*\gamma) = \frac12 \left( \frac2s -1 - \log \frac2s \right), \\ D(P*\gamma\|R*\gamma) = \frac12\left( 2 - 1 - \log2 \right).$$
But notice that the first expression above blows up as $s \to \infty,$ meaning that large $s$ yield a counterexample to the original statement.
We can also make this example nondegenrate, relying basically on the fact that $D$ blows up to $\infty$ smoothly as $\sigma_2 \to 0$. Indeed, parametrise the variances of $Q$ and $R$ as $s-1$ and $r$, where $s > 1, r < 1$. Then we have the conditions $$ \frac1{s-1} - \log \frac1{s-1} < \frac 1r - \log \frac1r\\ \frac2s - \log \frac2s > \frac2{r+1} - \log \frac2{r+1},$$ and we want to check if this has solutions $(r,s)$. For this, essentially any tiny $r,$ and somewhat large $s$ will work, but we can be more explicit: take $r = 0.5, s = 3$ (which I found with a CAS). We have $$\frac{1}{2} + \log(2) - (2 - \log(2)) = 2 \log(2) - 3/2 < -0.1\\ \frac23 - \log \frac23 - \left( \frac43 - \log \frac43 \right) = \log 2 - \frac23 > 0.02, $$ where I'm using the standard fact that $0.69 < \log(2) < 0.7.$