Let $k$ be a field of characteristic $p > 3$. I cannot find any example of ordinary isotrivial elliptic $k$-surface $E$ (i.e., elliptic $k(t)$-curve, where $t$ is a variable) whose Mordell-Weil group $E(k(t))$ contains the subgroup $\mathbb{Z} \times \mathbb{Z}/3$. In addition, I assume that the $j$-invariant of $E$ is different from $0, 1728$. If I required only $E(k(t)) \supset \mathbb{Z}/3$, we could just take the constant surface $E$ associated with an elliptic $k$-curve $E_0$ having an order $3$ point.

Is it possible to construct such an elliptic surface in principle? Curiously, there is a desired **non-isotrivial** elliptic surface according to the classical [article][1]. Also, for any $j$-invariant, we have an example of the isotrivial $E$ such that $E(k(t)) \supset \mathbb{Z} \times \mathbb{Z}/2$. Indeed, given an elliptic $k$-curve $E_0\!: y^2 = f(x)$, it is enough to consider its twist $E\!: f(t)y^2 = f(x)$. As is known (see, e.g., Proposition 3.1 of [Shioda's article][2]), the Mordell-Weil group of this $E$ is naturally isomorphic to $\mathrm{End}(E_0) \times E_0(k)[2]$.

If one can construct the required surface $E$ over a finite field $k = \mathbb{F}_{\!q}$ satisfying the additional condition $3 \mid (q-1)$, then the smooth fibers $E_t$ with $t \in \mathbb{F}_{\!q}$ are pairwise $\mathbb{F}_{\!q}$-isomorphic (not just $\overline{\mathbb{F}_{\!q}}$-isomorphic). This is a nice feature. Indeed, there are only two elliptic $\mathbb{F}_{\!q}$-curves of a given $j$-invariant $\neq 0, 1728$. Since, the order $\#E_t(\mathbb{F}_{\!q}) = q+1 - a$ is divisible by $3$ (with $a$ as the Frobenius trace), $\#E_t^T(\mathbb{F}_{\!q}) = q+1 + a$ is not, where $E_t^T$ is the unique non-trivial (quadratic) $\mathbb{F}_{\!q}$-twist of $E_t$. Otherwise, $3 \mid 2(q+1)$, that is, $3 \mid (q+1)$, which is a contradiction to our assumption.

  [1]: https://academic.oup.com/imrn/article-abstract/1993/4/113/674514?login=false
  [2]: https://www2.rikkyo.ac.jp/web/shioda/mathpapers2005-7/revcorrmwl.pdf