In the case where $\mathcal{C}$ is presentable, this is constructed in proposition 2.9 of
Robalo, Marco, $K$-theory and the bridge from motives to noncommutative motives, Adv. Math. 269, 399-550 (2015). ZBL1315.14030.
Note that if $\mathcal{C}$ is stable, so is $\mathcal{C}[X^{-1}]$, since $\mathcal{C}$ contains an inverse for $S^1$ and $\mathcal{C}\to \mathcal{C}[X^{-1}]$ is symmetric monoidal.
Furthermore, corollary 2.22 shows that if $X$ is a symmetric object (i.e. the cyclic permutation acts trivially on $X^{\otimes n}$ for some $n>1$), then $\mathcal{C}[X^{-1}]$ can be obtained as a category of spectrum objects, i.e. as the $\infty$-category $$\operatorname{Stab}_X(\mathcal{C}):=\operatorname{colim}\left(\mathcal{C}\xrightarrow{X\otimes - }\mathcal{C}\xrightarrow{X\otimes-}\cdots\right)\cong \operatorname{lim}\left(\mathcal{C}\xleftarrow{\operatorname{hom}(X,-)}\mathcal{C}\xleftarrow{\operatorname{hom}(X,-)}\cdots\right)$$ where the first colimit is taken in $\mathrm{Pr}^L$.
To see an example where the latter identification does not hold, take $\mathcal{C}=\mathrm{Sp}$ and $X=\mathbb{S}\oplus \mathbb{S}$. Then it can be shown that $\mathcal{C}[X^{-1}]=0$ (indeed every additive $\infty$-category where the functor $y\mapsto y\oplus y$ is fully faithful is 0), but $\operatorname{Stab}_X\mathcal{C}$ is not 0, since its K-theory is $K(\mathbb{S})[1/2]$. For more discussion of this particular example see this paper.