This is rather a question for mathoverflow. The canonical map $\pi^* : \mathrm{Pic}(X)\to\mathrm{Pic}(X\times\mathbb A^1)$ is an isomorphism when $X$ is normal, and there are counterexamples with $X$ local integral of dimension $1$ and of course non-normal (with no isomorphism between $\mathrm{Pic}(X)$ and $\mathrm{Pic}(X\times\mathbb A^1)$).
First suppose $X$ is normal. As $\pi^*$ is injective, it is enough to consider affine open subsets of $X$. Then descend to finitely generated $\mathbb Z$-algebras. Taking integral closure and because $\mathbb Z$ is excellent, we are reduced to the case $X$ noetherian and normal. Now apply EGA IV.21.4.11, page 360. Note that the same arguments apply for any non-empty open subset of $\mathbb A^1_\mathbb Z$, e.g., $\mathbb G_m$, instead of $\mathbb A^1$.
Now let us see a counterexample with $X$ non-normal. I will take the first integral non-normal example which comes to my mind: $X=\mathrm{Spec}(R)$ where $R$ is the local ring $$R=(k[u,v]/(u^2+v^3))_{(u,v)}$$
over a field $k$ of characteristic zero. Let $K=\mathrm{Frac}(R)$. As $X$ is local, $\mathrm{Pic}(X)$ is trivial. Denote by $Y:=X\times \mathbb A^1$. We want to show $\mathrm{Pic}(Y)$ is non-trivial.
Consider the polynomial
$$f=1+vT^2\in R[T].$$
It is chosen in such a way that $f$ is not irreducible in $K[T]$ :
$$f=(1+tT)(1-tT)=(v+uT)(1/v+(v/u)T),\quad t:=-u/v\in K$$
but is $f$ irreducible (I don't say prime) in $R[T]$ (I don't use explicitly this property, just to explaine where comes this $f$). Note that $Y$ is covered by the two affine open subsets $D(f)$ and the generic fiber $Y_K$. Let $L$ be the invertible sheaf on $Y$ given by
$$L({D(f)})=(v+uT)R[T]_f, \quad L({Y_K})=K[T].$$
This is well defined because $(v+uT)$ is an invertible element of $O_Y(D(f)\cap Y_K)=K[T]_f$.
Admit for a moment that
$(R[T]_f)^{\star}=R^{\star}f^{\mathbb Z}$.
If $L$ is free, then there exist $\omega=af^{r}\in (R[T]_f)^\star$ and $\lambda\in (K[T])^\star=K^\star$ such that $(v+uT)\omega=\lambda$. This is impossible by comparing the degrees of both sides in $K[T]$. So $\mathrm{Pic}(Y)$ is non-trivial. The above fact on the units of $R[T]_f$ is (with the new $f$) easy to see. But I can post my solution if you want.
After thinking about this for some time after today's lecture, I believe the answer to your question is "yes" and I'll attempt to give a detailed proof (which might be a bit long for a post, but here we go).
Let $M$ be any $\mathbb C$-scheme and $h^M=\operatorname{Hom}_{\mathrm{Sch}/\mathbb C}(-,M)$ its represented functor. Let $\eta\colon \operatorname{Pic}_{\mathbb A^1/\mathbb C}\rightarrow h^M$ be any natural transformation; we must show that $\eta_X\colon \operatorname{Pic}_{\mathbb A^1/\mathbb C}(X)\rightarrow h^M(X)$ has image a single point for all $\mathbb C$-schemes $X$.
Step 0. We reduce to the case where $X$ is affine. Consider any affine open cover $X=\bigcup U_i$ and the commutative diagram
The right vertical morphism is injective because $h^M$ is a Zariski-sheaf. Thus it suffices to show that each $\eta_{U_i}\colon \operatorname{Pic}_{\mathbb A^1/\mathbb C}(U_i)\rightarrow h^M(U_i)$ has image a single point.
Step 1. We do the case where $X$ is reduced (and affine). For a point $x\in X$ let $\kappa(x)$ denote its residue field. We first claim that $h^M(X)\rightarrow \prod_{x\in X} h^M(\operatorname{Spec}\kappa(x))$ is injective again. After passing to an affine cover of $X$ that is mapped into an affine cover of $M$ and using some simple arguments as in Step 0, this boils down to the following question: Let $A$ be a reduced ring and $f,g\colon B\rightarrow A$ two ring morphisms that agree after composition with $A\rightarrow \kappa(\mathfrak p)$ for all $\mathfrak p\in \operatorname{Spec} A$. Then $f=g$. Indeed, the difference $f-g$ must have image contained in $\bigcap_{\mathfrak p\in \operatorname{Spec} A}\mathfrak p=0$ since $A$ is reduced.
Now consider the diagram
and observe that the bottom left product is a single point because both $\operatorname{Pic}(\operatorname{Spec} \kappa(x)[t])$ and $\operatorname{Pic}(\operatorname{Spec} \kappa(x))$ are trivial for all $x\in X$ (since line bundles over a UFD are trivial). So injectivity of the right vertical arrow does the trick.
Step 2. We do the case where $X$ is noetherian (and affine). To this end we claim $\operatorname{Pic}(X)\cong \operatorname{Pic}(X^{\mathrm{red}})$ for every affine noetherian scheme $X$, which immediately reduces everything to Step 1 [Edit: Turns out it doesn't, but fortunately Nuno has found a beautiful fix.] (this also uses $(X\times \mathbb A^1)^{\mathrm{red}}\cong X^{\mathrm{red}}\times\mathbb A^1$ which follows from a simple inspection). To prove the claim, let $\mathcal J$ be the coherent sheaf on $X$ cutting out its reduction $X^{\mathrm{red}}$. Since $X$ is noetherian, $\mathcal J^n=0$ for some $n$. Doing induction on $n$ we may assume $\mathcal J^2=0$. Now consider the short exact sequence $$1\longrightarrow (1+\mathcal J)\longrightarrow \mathcal O_X^\times \longrightarrow \mathcal O_{X^{\mathrm{red}}}^\times\longrightarrow 1$$ of multiplicative sheaves on $X$ (or rather its underlying topological space, which is the same as of $X^{\mathrm{red}}$). Using $\mathcal J^2=0$, it's straightforward to check that $1+\mathcal J$ is isomorphic to $\mathcal J$ (as an additive sheaf of abelian groups on $X$). Since $X$ is affine, $H^1(X,\mathcal J)=0=H^2(X,\mathcal J)$, so the long exact cohomology sequence provides the desired isomorphism $\operatorname{Pic}(X)\cong H^1(X,\mathcal O_X^\times)\cong H^1(X^{\mathrm{red}}, \mathcal O_{X^{\mathrm{red}}}^\times)\cong \operatorname{Pic}(X^{\mathrm{red}})$.
Step 3. We consider general affine $\mathbb C$-schemes $X$. Write $X=\lim X_\alpha$ as a cofiltered limit of noetherian affine $\mathbb C$-schemes $X_\alpha$ with affine transition maps. Using [Stacks project, Tag 01ZR & Tag 0B8W], one obtains $\operatorname{Pic}(X)\cong\operatorname{colim}\operatorname{Pic}(X_\alpha)$. The same holds for $X\times \mathbb A^1\cong \lim(X_\alpha\times\mathbb A^1)$, so actually $\operatorname{Pic}_{\mathbb A^1/\mathbb C}(X)\cong\operatorname{colim}\operatorname{Pic}_{\mathbb A^1/\mathbb C}(X_\alpha)$. Now every $\eta_{X_\alpha}\colon \operatorname{Pic}_{\mathbb A^1/\mathbb C}(X_\alpha)\rightarrow h^M(X_\alpha)\rightarrow h^M(X)$ has image a single point by Step 2, hence the same must be true for $\eta_X$ by the fact that the colimit in question is filtered. This finishes the proof.
I believe the argument in Step 3 can also be used (with some care) to show $\operatorname{Pic}(X)\cong \operatorname{Pic}(X^{\mathrm{red}})$ for arbitrary affine schemes, which would give an alternative to Step 3.
Best Answer
All the material you need to attack this problem will be located in the chapter on divisors in your favorite algebraic geometry book. For instance, here's the relevant material from Hartshorne chapter II section 6:
Corollary 6.16: If $X$ is a noetherian, integral, separated locally factorial scheme, then there is a natural isomorphism $\operatorname{Cl} X\cong\operatorname{Pic} X$.
Proposition 6.5(c): Suppose $X$ is a noetherian integral separated scheme which is regular in codimension one. Let $Z$ be an irreducible closed subset of codimension one, and let $U$ be it's open complement. Then there is an exact sequence $$\Bbb Z\to\operatorname{Cl} X\to \operatorname{Cl} U\to 0$$ where $1\in\Bbb Z$ maps to $[Z]\in \operatorname{Cl} X$.
Combining these statements with the result that $\operatorname{Cl} \Bbb P^1_F=\Bbb Z$ generated by a closed $F$-rational point, we see that $\operatorname{Cl} \Bbb A^1_F=0$, and therefore every open subscheme also has vanishing class (and therefore Picard) group.
Alternately, one may use the homotopy invariance of the class/Picard group - see proposition 6.6 which states that if $X$ satisfies the assumptions of proposition 6.5 above, then $\operatorname{Cl} X\cong \operatorname{Cl} X\times \Bbb A^1$; or here for the statement that if $X$ is normal, then $\operatorname{Pic} X\cong \operatorname{Pic} X\times\Bbb A^1$. As the class/Picard group of $\operatorname{Spec} F$ is trivial, the class/Picard group of $\Bbb A^1_F$ is also trivial, and then we may continue on with proposition 6.5(c) as above.