Consider the complete theory of discrete linear orders without bounds $T$ in the language $\{<\}$. Now introduce a symbol for successor $S$ and add the axiom $y=S(x)\iff y>x\wedge \neg\exists z\ x<z<y$. Call the new expanded theory in the expanded language $T'$.
Is the deductive closure of $T'$ complete? I would say "yes", since every sentence with $S$ can be replaced by an equivalent sentence in the old theory which is complete.
Am I right? Is the same true (or false) for analogue extensions of complete theories?
Yes, the extended theory is complete. To see why, suppose that $\varphi$ is any statement in the expanded language, then we can recursively translate it to a statement in the language $<$, by tediously replacing each $S(x)$ with a bounded variable $\exists y(\ldots)$ satisfying the axiom defining $S$.
For example $\exists x\exists y(S(x)<S(y))$ would be translated to the formula: $$\exists x\exists y(\exists z(x<z\land\lnot\exists u(x<u<z)\land\exists w(y<w\land\lnot\exists u(y<u<w)\land z<w))).$$
Now since $T$ proves or disproves this translation of $\varphi$, it follows that $T'$ also proves it. And since $T'$ proves that we can replace $z$ by $S(x)$ and $w$ by $S(y)$, it follows that $T'$ proves or disproves $\varphi$.