I assume the base field is of characteristic zero below.
Let $f:X\to \mathbb{P}^1$ be a smooth proper morphism with fibres of dimension one. Note that the fibres of $f$ are geometrically connected by Stein factorization and the fact that $\mathbb{P}^1$ is simply connected.
Theorem. $f$ is isotrivial.
Proof. Let $g$ be the genus of the fibres. Clearly, if $g=0$, then all geometric fibres are isomorphic, so $f$ is isotrivial by definition. Next, assume that $g=1$. Then, as the moduli space of elliptic curves is affine, the moduli map associated to the Jacobian of $f$ is constant, so that $Jac(f)$ has constant $j$-invariant. It follows that $f$ is isotrivial. Finally, it is a theorem of Parshin that any genus $g>1$ curve over $\mathbb{P}^1$ is isotrivial. (More generally, any smooth proper morphism $X\to \mathbb{P}^1$ whose fibres are smooth proper connected varieties with ample canonical bundle is isotrivial by work of Kovács, Migliorini, Viehweg-Zuo.) Parshin's theorem follows from the "hyperbolicity" of the moduli stack of genus $g$ ($g>1$) curves. There are many notions of "hyperbolicity" lurking around there, and all of them imply the statement you want. QED
Thus, $f$ is isotrivial. Let $F$ be a fibre (over a closed point of $\mathbb{P}^1$).
If $g>1$, then the Isom-scheme $Isom(F\times \mathbb{P}^1, X)\to \mathbb{P}^1$ is finite etale and thus trivial. Thus, $f$ is trivial.
If $g=0$, then $f$ is a Brauer-Severi scheme. But all Brauer-Severi schemes over $\mathbb{P}^1$ are trivial, as the Brauer group of $\mathbb{P}^1$ is trivial.
If $g=1$, then probably the family $X\to \mathbb{P}^1$ is trivial. But I can't remember the correct argument at the moment.