$\DeclareMathOperator\Pic{Pic}$If $A$ is a ring, then we know that $\Pic(A)=\Pic(A_\text{red})$, but for a scheme $X$ it is false in general.
On the other hand, we have that $\Pic(X)=H^{1}_{et}(X,\mathbb{G}_m)$ and étale cohomology doesn't see the nilpotents, so there should be no difference of the right hand side if we replace $X$ by $X_\text{red}$.
How can we accomodate the two?
Let $f : X \to Y$ be a universal homeomorphism of schemes. Then as you noted above, the pullback functor on (small) étale sites
$$\begin{eqnarray} Y_{\text{ét}} &\to& X_{\text{ét}}\\ U &\mapsto& U \times_X Y \end{eqnarray} $$ is an equivalence of categories. In this case, the natural transformations $\text{id} \to f_\ast f^\ast$ and $f^\ast f_\ast \to \text{id}$ are isomorphisms. So now let $X = Y_{\text{red}}$. If $f^\ast \mathbf{G}_{m,Y} = \mathbf{G}_{m, Y_{\text{red}}}$, then it would follow for any etale open $U_{\text{red}}$ over $Y_{\text{red}}$ that $$\mathbf{G}_{m,{\text{red}}}(U_{\text{red}}) = \mathbf{G}_m(U), $$
where $U$ is the unique scheme étale over $Y$ that pulls back to $U_{\text{red}}$ over $Y_{\text{red}}$.
The following example shows this is already false for rings: Take $U= Y = \operatorname{Spec} k[\epsilon]/(\epsilon^2)$. Then $\mathbf{G}_m(Y) = k[\epsilon]^\times$ which is evidently not equal to $k^\times$. For instance, $1+\epsilon$ is a unit because $(1+ \epsilon)(1-\epsilon) = 1 - \epsilon^2 = 1$.
