Assume $(x_n)_{n\geq1}, (y_n)_{n\geq1}$ are sequences of elements of the Skorokhod space $\mathcal D = \mathcal D(\mathbb R)$ of cadlag functions with range in $\mathbb R$, endowed with the usual $J_1$ topology. Assume moreover that both $(x_n+y_n)_{n\geq1}$ and $(y_n)_{n\geq1}$ are tight families of random variables (for example, they converge to some limiting random variables).
Is it possible to deduce from this that $(x_n)_{n\geq1}$ is also a tight family? What makes it non obvious (to me) is that in general addition is not continuous in $\mathcal D$.
No, you can disprove this with degenerate random variables. Take any sequences $a_n \to a$ and $b_n \to b$ in $D_{\mathbb{R}}([0,\infty))$ with the property that $a_n - b_n$ is not relatively compact. Then if $y_n = b_n$ and $x_n = a_n - b_n$, you have a counterexample.
For example, suppose that $y_n$ is almost surely the constant sequence $y_n = 1_{\{x \geq 1\}}$ and let $x_n$ also be almost surely deterministic $x_n = 1_{\{x \geq 1 + \frac{1}{n}\}} - 1_{\{x \geq 1\}}$. The family $\{y_n\}$ is relatively compact because it is constant. $x_n + y_n = 1_{\{x \geq 1 + \frac{1}{n}\}}$ and you can check by hand that this converges to $1_{\{x \geq 1\}}$ in $D_{\mathbb{R}}([0,\infty))$. You can also check that $x_n$ does not converge (any limit would have to be zero on $[0,1)$ and $(1,\infty)$).
As a comment, if you have tightness then you have subsequential limits in distribution. So long as you are working on a Polish space, Skorokhod's representation theorem says that you can assume without loss of generality that these limits are almost sure. Almost sure convergence means that pointwise almost everywhere you need to have relative compactness in the topology of the space, so typically you really can't hope for tightness in situations when you don't have relative compactness.