= Solution
There are two common normalizations. I will define both explicitly, because the numerical conclusion in the question uses the unreduced one.
Set $\delta=-A^2-A^{-2}$. The reduced <Kauffman bracket> of a nonempty <link diagram> is
$$
\langle D\rangle_{\mathrm{red}}
=\sum_s A^{a(s)-b(s)}\delta^{|s|-1},
$$
where $s$ runs over all bracket smoothings, $a(s),b(s)$ count $A$- and $B$-smoothings, and $|s|$ is the number of resulting circles. Locally,
$$
\langle\text{crossing}\rangle
=A\langle A\text{-smoothing}\rangle+A^{-1}\langle B\text{-smoothing}\rangle.
$$
Fix the usual <Kauffman bracket> convention in which the $A$-smoothing at a positive braid crossing is its <oriented smoothing>; reflecting a crossing exchanges $A$ and $A^{-1}$. The <unknot> has reduced bracket $1$, and adjoining another circle multiplies the bracket by $\delta$. The unreduced <Kauffman bracket> uses $\delta^{|s|}$ instead and assigns the empty diagram $1$. Thus for a nonempty diagram
$$
\langle D\rangle_{\mathrm{un}}=\delta\langle D\rangle_{\mathrm{red}}.
$$
For completeness, the bracket calculations for all three <Reidemeister moves> can be carried out in the <Temperley-Lieb diagram algebra>. Let $I$ be the identity two-strand <tangle>, and let $e$ be the cap-cup <tangle>. Composition gives $e^2=\delta e$. The two crossings correspond to
$$
R=A I+A^{-1}e,\qquad R^{-1}=A^{-1}I+A e.
$$
Consequently
$$
RR^{-1}=I+(A^2+A^{-2}+\delta)e=I,
$$
which proves invariance under the second <Reidemeister move>. On three strands, the cap-cup diagrams satisfy $e_1e_2e_1=e_1$, $e_2e_1e_2=e_2$, and $e_i^2=\delta e_i$. Expand the two sides of the third move. Their difference is
$$
R_1R_2R_1-R_2R_1R_2
=(A+A^{-1}\delta+A^{-3})(e_1-e_2)=0.
$$
Thus the third <Reidemeister move> also preserves either bracket normalization. Reflected forms of these moves follow by replacing $A$ with $A^{-1}$.
A positive curl contributes $A\delta+A^{-1}=-A^3$, and a negative curl contributes $A^{-1}\delta+A=-A^{-3}$. The <writhe of a link diagram> $w(D)$ changes by $+1$ or $-1$ in exactly these cases. Therefore the corrected bracket
$$
f_D(A)=(-A^3)^{-w(D)}\langle D\rangle
$$
is invariant under the first <Reidemeister move> as well; the second and third moves leave the <writhe of a link diagram> unchanged. Substituting $t=A^{-4}$ defines a <Jones polynomial>. For multiple components half-integer powers of $t$ may occur. The reduced version $V_L^{\mathrm{red}}$ has $V_U^{\mathrm{red}}=1$, and the <Unreduced Jones polynomial> is
$$
V_L^{\mathrm{un}}(t)=-(t^{1/2}+t^{-1/2})V_L^{\mathrm{red}}(t).
$$
At $A=1$, switching a crossing does not change the bracket: both smoothing coefficients are $1$. It changes the <writhe of a link diagram> by $2$, so the correction factor $(-1)^{-w}$ is unchanged. Hence either <Jones polynomial> at $t=1$ is unchanged by a <crossing change>. By switching crossings, any $l$-component <link> can be made an <unlink>. Its crossing-free diagram has $l$ circles and zero <writhe of a link diagram>, giving
$$
\boxed{V_L^{\mathrm{red}}(1)=(-2)^{l-1},\qquad V_L^{\mathrm{un}}(1)=(-2)^l.}
$$
Thus \b[the printed conclusion $|V_L(1)|=2^l$ holds for the unreduced normalization]. For the reduced <Jones polynomial>, the correct conclusion is $\boxed{|V_L^{\mathrm{red}}(1)|=2^{l-1}}$; the <unknot> alone already rules out $2^l$ in that convention.
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