You must have meant to assume $s \in B$, since $s$ lies in the identity component of its centralizer (in the ambient connected semisimple group that you did not name) and so lies in every Borel subgroup of that identity component.

Consequently, your question can only have an affirmative answer if $s \in B_s$, which entails assume $s \in B$. So let's just forget about $P$ and assume $s \in B$.

And as ya-tayr notes, you need to fix up some connectedness issues (though they are not a problem when the original connected reductive group is semisimple and *simply connected*, which ${\rm{PGL}}_2$ in ya-tayr's example is not; I address this below).

The answer is "yes" if you fix up the connectedness issues, but the question as formulated has some serious lack-of-clarity issues for the following reason. This is a question that is intimately tied up with the structure of an underlying *algebraic group* (i.e., it is not a question of pure group theory), or at least of an abstract group equipped with a Tits system (which you also have not mentioned). The formulation is abusing terminology by speaking of finite groups of rational points "as if" they are linear algebraic groups, which they are not. This creates confusion with respect to connectedness aspects (as those are not well-suited to discussing via the language of *finite* groups of rational points over a finite field).

To be clearer, one should give a name to the (finite) ground field, let's call it $k$, and introduce the actual linear algebraic group with its own name, so let's write $\mathbf{G}$ to denote a connected reductive group over $k$. Then finally we have the ordinary group $G = \mathbf{G}(k)$.

We're given a Borel $k$-subgroup $\mathbf{B}$ of $\mathbf{G}$. We define $B = \mathbf{B}(k)$. Of course, we know by the Borel--Tits structure theory that $B$ as a subgroup of $G$ always determines $\mathbf{B} \subset \mathbf{G}$ uniquely (even if $k$ is finite), so it is only a felony (rather than a capital crime) to call $B$ a "Borel subgroup". But it is important/useful to maintain a clear distinction between the algebraic groups and their sets of rational points (especially over finite fields), so it is best to bring $\mathbf{B}$ out of the shadows.

Anyway, below I will solve what I think is the problem you have in mind, with the connectedness aspects handled correctly. I will also allow the ground field $k$ to be arbitrary.

We choose $s \in B$ that is semisimple (a concept whose definition involves $\mathbf{G}$ or $\mathbf{B}$, not just $G$ or $B$, though one can "cheat" over finite fields and give a definition in terms of the order of $s$ relative to the characteristic of $k$, but we won't do that).

Working with the individual element $s$ is the "wrong" thing to do. Let's instead consider the Zariski closure $\mathbf{M}$ in $\mathbf{G}$ of the subgroup of $G$ generated by $s$. Clearly $\mathbf{M}$ is a smooth closed $k$-subgroup of $\mathbf{B}$ "of multiplicative type": it is commutative with torus identity component and component group whose order is not divisible by ${\rm{char}}(k)$.

Of course, if $k$ is finite then this may look a bit silly, but the arguments below will work over any field whatsoever, and over an infinite field there is some content to thinking about $\mathbf{M}$ when $s$ has infinite order.

Clearly $s$-centralizers and $\mathbf{M}$-centralizers amount to the same thing, but $\mathbf{M}$ has more direct contact with algebraic geometry (unlike the poor isolated element $s$). Feel free to ignore this and instead use "$s$" everywhere below, but I find it more natural to treat tori and semisimple elements by a unified argument.

The key thing about $\mathbf{M}$ that works its magic is that the linear representation theory of $\mathbf{M}$ is *completely reducible* (i.e., $\mathbf{M}$ is "linearly reductive"). In what follows, $\mathbf{M}$ will be allowed to be any closed $k$-subgroup scheme of multiplicative type in $\mathbf{G}$ (e.g., I allow $\mu_p$ in characteristic $p > 0$) -- if you don't know what that means in the absence of smoothness, feel free to assume it is smooth (or even constant finite cyclic of order not divisible by ${\rm{char}}(k)$) if you prefer.

The main technical input we need is that for any *smooth* separated $k$-scheme $X$ equipped with an action by $\mathbf{M}$, the subfunctor $X^{\mathbf{M}}$ of $\mathbf{M}$-fixed points in $X$ (i.e., for $k$-algebra $A$, its set of $A$-points consists of $x \in X(A)$ that are invariant under the action of $\mathbf{M}_A$ on $X_A$) is represented by a *smooth* closed subscheme of $X$. This smoothness is proved via the infinitesimal criterion, ultimately boiling down to the linear reductivity of $\mathbf{M}$. In the cases that come up below, you will know this fact by other means, so feel free to ignore that generality if you wish.

Let's consider the (scheme-theoretic) centralizer $Z_{\mathbf{G}}(\mathbf{M})$, which is a *smooth* closed $k$-subgroup of $\mathbf{G}$ whose set of $k$-points is $C_G(s)$. There is a good reason for using the scheme-theoretic definition of this centralizer (even though it turns out to be smooth), namely that it tells us that the tangent space at the identity for this centralizer is "as expected", namely the $\mathbf{M}$-invariant elements inside $\mathfrak{g} := {\rm{Lie}}(\mathbf{G})$ under the adjoint representation of $\mathbf{G}$.

Since $\mathbf{M}$ is of multiplicative type, we know that the identity component $Z_{\mathbf{G}}(\mathbf{M})^0$ is reductive. (Again, this is well-known for the $\mathbf{M}$ of initial interest.) By a theorem of Steinberg, this centralizer is connected if $\mathbf{G}$ is semisimple and simply connected and $\mathbf{M}$ is smooth with cyclic identity component (as in the case of initial interest with $\mathbf{M}$ generated by $s$), but I have no idea if simple connectedness holds in the cases you care about, so I will be attentive to the connectedness in what follows.

For any smooth closed $k$-subgroup $\mathbf{H}$ of $\mathbf{G}$ normalized by
$\mathbf{M}$, the scheme-theoretic intersection $$\mathbf{H}_{\mathbf{M}} := \mathbf{H} \cap Z_{\mathbf{G}}(\mathbf{M})^0$$ is *smooth* (but perhaps disconnected, even if $\mathbf{H}$ is connected?). To see this, contemplate the centralizer in the smooth affine $k$-group $\mathbf{H} \ltimes \mathbf{G}$ under the conjugation action of the multiplicative type subgroup $1 \ltimes \mathbf{M}$

The reason I prefer the scheme-theoretic viewpoint on the intersection is because it provides the "expected" Lie algebra inside $\mathfrak{g}$ right away (though we had to do work to know the intersection is smooth). Notationwise, note that $\mathbf{G}_{\mathbf{M}} := Z_{\mathbf{G}}(\mathbf{M})^0$.

Taking $\mathbf{H} = \mathbf{B}$, we conclude that $\mathbf{B}_{\mathbf{M}}$ is a smooth closed $k$-subgroup of the connected reductive $\mathbf{G}_{\mathbf{M}}$ (with set of $k$-points essentially equal to $B_s$ as you defined it when $\mathbf{M}$ is the Zariski closure of the subgroup of $G$ generated by $s$, except that you overlooked the issue of $\mathbf{G}_{\mathbf{M}}$ possibly being disconnected; again, if $\mathbf{G}$ has been semisimple and simply connected then you would have been fine in the cases of interest, thanks to Steinberg's theorem). Clearly $\mathbf{B}_{\mathbf{M}}$ is solvable. Note that we definitely did not manually insert "identity component" into the definition of $\mathbf{B}_{\mathbf{M}}$; at this point we are ignorant about its connectedness status.

Finally, we come to the question I think you meant to ask: is $\mathbf{B}_{\mathbf{M}}$ a Borel subgroup of $\mathbf{G}_{\mathbf{M}}$? Let's show that the answer is affirmative. To prove this, it suffices to show that the smooth quasi-projective quotient scheme $\mathbf{G}_{\mathbf{M}}/\mathbf{B}_{\mathbf{M}}$ is proper (equivalently, projective) over $k$ (as then $\mathbf{B}_{\mathbf{M}}$ would be a parabolic $k$-subgroup, hence also connected by a theorem of Chevalley).

To be more precise, consider the natural map $f:\mathbf{G}_{\mathbf{M}} \rightarrow (\mathbf{G}/\mathbf{B})^{\mathbf{M}}$ into the scheme of $\mathbf{M}$-fixed points in $\mathbf{G}/\mathbf{B}$ (using the left-translation action). This fixed-point scheme is *smooth* precisely because $\mathbf{M}$ is linearly reductive. Once again, the advantage of using the scheme-theoretic definition is that the knowledge of its tangent space at the identity point comes for free: the $\mathbf{M}$-fixed locus in the tangent space $\mathfrak{g}/\mathfrak{b}$ of $\mathbf{G}/\mathbf{B}$ at the identity point. (This will be used at the end.)

The map $f$ is really the orbit map at the identity coset for the left action of $\mathbf{G}_{\mathbf{M}}$ on $(\mathbf{G}/\mathbf{B})^{\mathbf{M}}$, and the isotropy subgroup for the identity coset under this action is exactly $\mathbf{B}_{\mathbf{M}}$ due to the definitions. Hence, we get a locally closed immersion
$$j:\mathbf{G}_{\mathbf{M}}/\mathbf{B}_{\mathbf{M}} \rightarrow
(\mathbf{G}/\mathbf{B})^{\mathbf{M}}$$
into a target that is *complete* (since it is closed in $\mathbf{G}/\mathbf{B}$). Thus, we win if $j$ has closed image.

Even better: for *any* smooth closed $k$-subgroup $\mathbf{H}$ of $\mathbf{G}$ containing $\mathbf{M}$, we claim that the $\mathbf{G}_{\mathbf{M}}$-orbits on $(\mathbf{G}/\mathbf{H})^{\mathbf{M}}$ are all *open*. Since orbits are pairwise disjoint, it would follow that there are only finitely many orbits and they are thus also all closed.
This openness assertion is part of the proof of Proposition 11.15 in Borel's textbook on algebraic groups in the case that $\mathbf{M}$ is a torus (except that Borel assumes $\mathbf{H}$ is a Borel subgroup). The argument which follows is a variation which works for any $\mathbf{M}$ as above (since for the case of initial interest, tori will certainly be insufficient!).

We may assume $k$ is *algebraically closed* (aha, good that we didn't demand $k$ to be finite in the preceding discussion), so the problem is to show that for any $x \in (\mathbf{G}/\mathbf{H})^{\mathbf{M}}(k)$, the $\mathbf{G}^{\mathbf{M}}$-orbit through $x$ is open in $(\mathbf{G}/\mathbf{H})^{\mathbf{M}}$. Choose $g \in \mathbf{G}(k)$ lifting $x$. Since $\mathbf{H}$ contains $\mathbf{M}$, the $\mathbf{M}$-invariance of $x$ says exactly that $g^{-1}\mathbf{M}g \subset \mathbf{H}$. Moreover, left multiplication on $\mathbf{G}/\mathbf{H}$ by $g^{-1}$ carries the $\mathbf{M}$-fixed point scheme to the $g^{-1}\mathbf{M}g$-fixed point scheme, and it turns the orbit map through $x$ into the orbit map through $1$ at the cost of replacing $\mathbf{M}$ with its $g^{-1}$-conjugate (which also lies inside $\mathbf{H}$!). So it suffices to prove the openness property for the orbit map through 1.

[It might seem we have just come back to the original problem of the orbit through 1, but that is not so: only by considering the totality of *all* $\mathbf{G}^{\mathbf{M}}$-orbits does it suffice to prove openness in our aim to really proved closedness.]

OK, so now it remains to show (after renaming $g^{-1}\mathbf{M}g$ as $\mathbf{M}$) that the natural map $\mathbf{G}_{\mathbf{M}} \rightarrow (\mathbf{G}/\mathbf{H})^{\mathbf{M}}$ has open image. Even better, it is a smooth morphism. By homogeneity, to prove such smoothness it suffices to do so at a single point. We choose the identity point. And by *smoothness* of the source and target, it suffices to prove surjectivity of the map on tangent spaces. Ah, *now* we finally exploit that our scheme-theoretic constructions have the "expected" tangent spaces, so the map on tangent spaces is exactly the natural map
$$\mathfrak{g}^{\mathbf{M}} \rightarrow (\mathbf{g}/\mathfrak{h})^{\mathbf{M}}.$$
But the formation of $\mathbf{M}$-invariants is (right-)exact precisely because of the linear reductivity of $\mathbf{M}$, so we win. Voila.

QED