Hi stupid_question_bot,
Unfortunately you seem to need some more assumptions for an easy proof, in particular properness would make this very easy, in general given a flat proper scheme with geometrically normal fibers one can show that the number of (geometric) components of the fibers is locally constant on the base, which would answer your question.
(EDIT: To be clear, the following is not a counter-example to the specific statement in the question, that comes later in this answer. I was just trying to point out that the proof would need some geometric input since it is false when the base is not normal.)
The counter example that I have in mind is as follows, take $\bar{X}$ to be the nodal cubic over $\mathbb{Z}_p$ ($\mathbb{P}^1$ glued together at two $\mathbb{Z}_p$ points: say $0, 1$ in a standard affine chart), let $\bar{Y} \to \bar{X}$ be a connected finite etale cover corresopnding to a nontrivial element of the geometric fundamental group of $\bar{X}$ (for definiteness, take the double cover given by two $\mathbb{P}^1$'s glued into a bigon and for safety let $p \neq 2$). Now let $X$ be the complement of the node in the special fiber, and let $Y$ be the pullback. Clearly while the generic fiber of $Y$ is connected the special fiber is not by inspection.
You can now complain: "oh but your $X$ is not an snc complement in a smooth scheme." In this case I was unable to say anything useful, except for that some results in SGA imply that this would be true if the cover $Y$ is tamely ramified over the snc divisor. Hope this example is helpful though, as it shows that the strong statement you made about connectivity of special fibers is not some total triviality.
EDIT: Update, bad news: there are even worse examples to be had here. Let $X$ be $\mathbb{A}^1_{\mathbb{F}_p[[t]]}$, then consider $Y$ the Artin-Schreier cover of $X$ given by the equation $Y^p - Y = x \cdot t$, then the special fiber of this etale cover splits but generically it defines a Galois Artin-Schreier cover.
I am going to give an identification of $DD(M)(S)$ and $M_S(S)$.
Using the identification $D(M)=\mathrm{Spec}\mathcal{O_S}[M]$ and $\mathbb{G}_m=D(\mathbb{Z})$, we have
$$DD(M)(S)=\mathrm{Hom}_{\mathcal{O}_S-\mathrm{Hopf-alg}}(\bigoplus_{i\in \mathbb{Z}}e_i\mathcal{O}_S,\bigoplus_{i\in M}e_i\mathcal{O}_S)$$
i.e. every element is determined by a map $\phi$ of $\mathcal{O}_S$-modules s.t. the following commutative diagram holds
$$\require{AMScd}\begin{CD}
\bigoplus_{i\in \mathbb{Z}}e_i\mathcal{O}_S @>\phi>> \bigoplus_{i\in M}e_i\mathcal{O}_S\\
@VV{e_i\mapsto e_i\otimes e_i}V @VV{e_i\mapsto e_i\otimes e_i}V\\
\bigoplus_{j,k\in \mathbb{Z}}e_{j}\mathcal{O}_S\otimes e_k\mathcal{O}_S @>\phi\otimes \phi>> \bigoplus_{a,b\in M}e_a\mathcal{O}_S\otimes e_b \mathcal{O}_S
\end{CD}$$
It's equivalent to the data $\mu :\mathcal O_S \cong e_1\mathcal{O}_S\stackrel{\phi}{\to}\bigoplus_{i\in M}e_i\mathcal{O}_S$ (written $\mu=\sum_i e_i \mu_i$ where $\mu_i:\mathcal{O}_S\stackrel{\mu}{\to}\bigoplus_{j\in M}e_j\mathcal{O}_S\to e_i\mathcal{O}_S\cong \mathcal{O}_S$) s.t. the diagram commutes
$$\require{AMScd}\begin{CD}
\mathcal{O}_S @>\mu>> \bigoplus_{i\in M}e_i\mathcal{O}_S\\
@VV{\mathrm{id}}V @VV{e_i\mapsto e_i\otimes e_i}V\\
\mathcal{O}_S @>{\mu\otimes\mu}>> \bigoplus_{a,b\in M}e_a\mathcal{O}_S\otimes e_b \mathcal{O}_S
\end{CD}$$
and $\sum_i \mu_i=\mathrm{id}$.
It's equivalent to the condition that $(\mathcal{O}_S\otimes \mathcal{O}_S\stackrel{\Delta}{\to}\mathcal{O}_S)\circ(\mu_a \otimes \mu_b)=\delta_{ab}\mu_a$ and $\sum_i \mu_i=\mathrm{id}$.
Over any affine open, we have $(\mathcal{O}_S\otimes \mathcal{O}_S\stackrel{\Delta}{\to}\mathcal{O}_S)\circ(\mu_a \otimes \mu_b)=\mu_a \circ \mu_b$. So it's equivalent to the condition that $\mu_a \circ \mu_b =\delta_{ab}\mu_a$ and $\sum_i \mu_i =\mathrm{id}$.
Claim: It's equivalent to the data {$(U_i)_{i\in M}$ is a disjoint open cover of $S$}.
Proof of the claim. Given a disjoint open cover $(U_i)_{i\in M}$ of $S$, then each $U_i^c=\bigcup_{j\neq i}U_j$ is open, there exist an unique element $c_i\in \mathcal{O}_S(S)$ s.t. $c_i|_{U_i}=1$ and $c_i|_{U_i^c}=0$ using axioms of sheaf. With each $c_i$ we can associate $\mu_i:\mathcal{O}_S\to \mathcal{O}_S,u\mapsto c_i u$. It's not hard to see that $\mu_a \circ \mu_b=\delta_{ab}\mu_a$. As $\sum_i c_i=1$, we have $\sum_i \mu_i=\mathrm{id}$.
Reversely for each $s\in S$, $\mu_{i,s}:\mathcal{O}_{S,s}\to\mathcal{O}_{S,s}$ is completed determined by $c_{i,s}=\mu_{i,s}(1)$. We know that $c_{i,s}c_{j,s}=\delta_{ij}c_{i,s}$ and $\sum_{i}c_{i,s}=1$. So each $c_{i,s}$ is idempotent ($x^2=x$), but the only idempotent elements in a local ring is $0$ and $1$. So exactly one of $(c_{i,s})_{i\in M}$ is 1.
For any $i$, denote $U_{i}$ as the subset consisting of $s$ s.t. $\mu_{i,s}=\mathrm{id}$. Then $(U_{i})_{i\in M}$ is disjoint. If $s\in U_i$, then $\mu_{i,s}=\mathrm{id}$, clearly it extends to an open neighborhood of $s$, see tag 01CP. Thus $(U_{i})_{i\in M}$ is disjoint open cover of $S$. The result follows.$\square$
Clearly the data {$(U_i)_{i\in M}$ is a disjoint open cover of $S$} is equivalent to the data {locally constant function $S\to M$}. The result follows.
Best Answer
To make the answer self-consistent, I recall some notions. The direct answer is the bold part.
When $S=\mathrm{Spec} k$ for a field $k$ and $G$ is just an $S$-group scheme, by [SGA3 I$_{\text{new}}$, VI$_{\text{A}}$, 2.6.5], there is a unique subgroup scheme $G^0\subset G$ whose underlying space is the irreducible component of $G$ containing the neutral section.
When $S$ is local Artinian and $G$ is locally of finite type over $S$, then $G^0$ also exists ([SGA3 I$_{\text{new}}$, VI$_{\text{A}}$, 2.3]) as a subgroup scheme. If $G$ is further assumed to be $S$-flat, then the fpqc sheaf quotient $G/G^0$ is an $S$-etale group scheme; if $S$ is further assumed to be an algebraically closed field, then the set $\mathcal{C}$ of connected components is a $G/G^0$-torsor, namely $\mathcal{C}\simeq G/G^0$, so we obtain the desired identification.
However, when $k$ is not separably closed, then by the above case, $\mathcal{C}$ will correspond to the set of $\mathrm{Gal}(k^{\mathrm{sep}}/k)$-orbits of $(G/G^0)(k^{\mathrm{sep}})$. For instance, $\mu_{3,\mathbf{Q}}=\mathbf{Q}[X]/(X^3-1)$ is a finite etale $\mathbf{Q}$-group scheme, but it has only two connected components: the neutral component $c_1$ and the other $c_2$ obtained by gluing $\zeta_3$ and $\zeta_3^{-1}$ together via the Galois action. The point is that there is no group structure on $\{c_1,c_2\}$ compatible with the group structure of $\mu_{3,\mathbf{Q}}$. (You can always endow a finite set with a group structure, but it will make the question meaningless: the compatibility is very important)
The following recollection is not directly relevant to the question, but I write here just for someone's convenience, especially who concerns about the existence of $G^0$.
For a scheme $S$ and an $S$-group scheme $G$, denote by $\underline{G^0}$ the subset of $G$ as the union of $G^0_s$ for all $s\in S$, where $G^0_s$ is the connected component of $G_s$ containing the neutral section. By [SGA3 I$_{\text{new}}$, VI$_{\text{B}}$, 3.4], if $\underline{G^0}\subset G$ is an open subset, then the subfunctor $$G^0\colon S^\prime/S\mapsto \{u\in G(S^\prime)|u(S^\prime)\subset \underline{G^0}\}$$ is representable by an $S$-subgroup scheme whose underlying space is $\underline{G^0}$.
Indeed, when $G$ is an $S$-flat finitely presented group scheme, then [EGAIV$_3$, 15.6.5] implies that $\underline{G^0}$ is open in $G$. In particular, $G^0$ is an open $S$-group scheme of $G$.
When $S$ is Noetherian of finite Krull dimension and $G$ is commutative smooth of finite type over $S$, then $G^0$ exists as an $S$-smooth finite type group and $G/G^0$ is an etale group algebraic space, see the paper of Giuseppe Ancona, Annette Huber and Simon Pepin Lehalleur, On the relative motive of a commutative group scheme.