I already know that the product of an elliptic curve and any curve of genus greater that 2 will give rise to a surface of Kodaira dimension 1.
Here is one strategy I have been suggested. Take a rational surface over the projective line that is a genus 1 fibration. For example the blow up of the base locus of a pencil of cubics. Then base change this to a curve of degree greater that 2 through a finite morphism. I am however unable to show that the resulting construction has Kodaira dimension 1 or that it is not the product of 2 curves.
To repeat the setup of the question, let $f \colon S \rightarrow \mathbb P^1$ be a rational elliptic surface, let $g \colon C \rightarrow \mathbb P^1$ be a finite cover from a curve of genus at least 2, and let $\pi \colon X \rightarrow C$ be the base change of $f$ along $g$. So $\pi$ is an elliptic fibration with a nonempty set of singular fibres.
First we show that the Kodaira dimension of $X$ is $\kappa(X)=1$. Note that if $E$ is any smooth fibre of $\pi$, then adjunction gives that the restriction $(K_X)_{|E}$ is trivial. Since such curves $E$ cover a dense open set of $X$, this shows that $\kappa(X) \neq 2$. (Unfortunately, easy arguments like this don't seem sufficient to finish the proof.) On the other hand, by Iitaka's Conjecture $C_{2,1}$ (Barth--Peters--Hulek--van de Ven Theorem III.18.4) we have $\kappa(X) \geq \kappa(C) + \kappa(E) = 1$, so we conclude that $\kappa(X)=1$.
Now we show that $X$ is not a product of curves. Suppose $X=C_1 \times C_2$ where the $C_i$ are smooth curves. Then $\kappa(X) = \kappa(C_1)+\kappa(C_2)$. Since $\kappa(X)=1$ the only possibility is that one of the curves, $C_1$ say, is an elliptic curve. Considering any fibre $C_1 \times {p}$, the image $\pi(C_1 \times \{p\})$ must be a point, since $C_1$ can't dominate a curve of genus $\geq 2$ such as $C$. So every fibre of $X \rightarrow C_2$ is contained in a fibre of $\pi$. But this is impossible, for example because some of the fibres of $\pi$ are singular.
Footnote: the OP also asked in comments how to see that $\chi(O_X)$ is positive, as is required by the canonical bundle formula that I initially invoked in my proof. BHPvdV Proposition V.12.2 shows that $\chi(O_X) = \operatorname{deg}(\pi_* K_{X/C}))$, and the latter is non-negative by Theorem III.18.2 of op. cit. But the proof of this theorem is non-trivial and I can't really say any more about it than "read the book".