I believe the following works (although it needs to be double-checked). For each of $i=1,2$, let $\nu_i:S_i \to P_i$ be the blowing up of a projective plane $P_i$, isomorphic to $\mathbb{P}^2$, at the 9 base points of a general pencil of plane cubics. Let $B_i$, isomorphic to $\mathbb{P}^1$, be the parameter space for this pencil of plane cubics, and let $\pi_i:S_i \to B_i$ be the corresponding elliptic fibration. Thus there is a projection $$\pi_1\times \pi_2: S_1 \times S_2 \to B_1 \times B_1,$$ whose geometric generic fiber is isomorphic to an Abelian surface (in fact $\pi_i$ has a section, so also $\pi_1\times \pi_2$ has a section). By my computation, the dualizing sheaf of $S_1\times S_2$ is the pullback $(\pi_1\times \pi_2)^*[\mathcal{O}_{B_1}(-1)\otimes \mathcal{O}_{B_2}(-1)]$, where $\mathcal{O}_{B_i}(+1)$ is the unique ample generator of the Picard group of $B_i$ (usual Serre twisting sheaf).
Now let $f:B_2\to B_1$ be a general choice of isomorphism. The graph is $g:B_2\to B_1\times B_2$. The normal bundle of the divisor $g(B_2)$ is $[\mathcal{O}_{B_1}(-1)\otimes \mathcal{O}_{B_2}(-1)]$. Thus, by adjunction, the inverse image $X$ of this divisor under $\pi_1\times \pi_2$ is a divisor in $S_1\times S_2$ that has trivial dualizing sheaf. By Bertini's theorem, for general choice of $f$, $X$ is smooth. Thus $X$ is a smooth, projective $3$fold with trivial dualizing sheaf. Also the fibration $\pi:X\to g(B_2)$ is an Abelian fibration over $B_2$ (I will identify $g(B_2)$ with $B_2$).
What about the K3 fibration? Let $x$ be any of the $9$ base points of the pencil of plane cubics on $S_1$. Let $\widetilde{P}_1 \to P_1$ be the blowing up along $x$. Linear projection away from $x$ defines a morphism $\rho':\widetilde{P}^1 \to \Lambda$, where $\Lambda$ is isomorphic to $\mathbb{P}^1$ and $\widetilde{P}^1$ is a $\mathbb{P}^1$-bundle over $\Lambda$. Since $S_1$ is the blowing up of $P_1$ along $9$ points that includes $x$, also $S_1$ is a blowing up of $\widetilde{P}_1$, $\mu:S_1\to \widetilde{P}_1$, at the transforms of the remaining $8$ points. Denote by $\rho:S_1 \to \Lambda$ the composition of $\mu$ and $\rho'$. The general fiber of $\rho$ is isomorphic to $\mathbb{P}^1$, but there are $8$ reducible fibers (isomorphic to a union of two copies of $\mathbb{P}^1$ intersecting at a single ordinary double point).
Consider the projection $\rho_1:S_1\to S_2 \to \Lambda$ that is the composition of the projection $\text{pr}_1:S_1\times S_2 \to S_1$ with $\rho$. Denote by $\rho_X:X\to \Lambda$ the restriction of $\rho_1$ to $X$. The claim is that a general fiber of $\rho_X$ is an elliptically fibered K3 surface.
The fiber $F$ of $\rho$ over a general point $t$ of $\Lambda$ is isomorphic to $\mathbb{P}^1$. Moreover, the projection $\pi_1|_F:F\to B_1$ is a degree $2$ cover of $B_1$ branched over $2$ general points. Thus the fiber $X_t$ of $\rho_X$ over $t$ is the fiber product of $\pi_1|_F:F\to B_1$ and $f\circ \pi_2:S_2\to B_1$. For general choice of $t$, $X_t$ is a smooth surface. Of course $S_2\to B_1$ has relative canonical bundle isomorphic to the pullback of $\mathcal{O}_{B_1}(+1)$. Thus $X_t \to F$ has relative canonical isomorphic to the pullback of $\mathcal{O}_{B_1}(+1)$, which is the same as the pullback of $\mathcal{O}_F(+2)$. Since $\omega_F$ is isomorphic to $\mathcal{O}_F(-2)$, this means that $X_t$ has trivial dualizing sheaf. So $X_t\to F$ is an elliptic fibration over $\mathbb{P}^1$ with trivial dualizing sheaf. Thus $X_t$ is an elliptic K3 surface. Thus $X\to \Lambda$ is a fibration over $\mathbb{P}^1$ by K3 surfaces.