First of all, note that your statement
Let $(M,g)$ be a 4-dimensional Riemannian manifold which is Kähler with respect to two independent complex structures $J_1, J_2$. Then $(M,g)$ admits a hyperkähler structure, in particular it is Ricci-flat.
is wrong. You can have a Kähler surface (which is a 4d Riemannian manifold) with precisely two Kähler forms if they have different orientation, i.e. belong to different bundles $\Lambda_{+}^2 TM$ and $\Lambda_{-}^2 TM$. Such geometries are called bi-Kähler. As an example, think of a direct product of two Riemannian surfaces with coordinates $(x_1, \phi_1)$ and $(x_2, \phi_2)$, each with its own axisymmetry $\partial_{\phi_1}$, $\partial_{\phi_2}$
$$
g = h_1(x_1) d x_1^2 + \frac{d \phi_1^2}{h_1(x_1)} + h_2(x_2) d x_2^2 + \frac{d \phi_2^2}{h_2(x_2)} \\
J_1 = d x_1 \wedge d \phi_1 + d x_2 \wedge d\phi_2\\
J_2 = d x_1 \wedge d \phi_1 - d x_2 \wedge d\phi_2\\
$$
Notice that one of the Kähler forms is self-dual, while the other is anti-self-dual. The geometry is hyperKähler only upon some differential conditions on the functions $h_1, h_2$.
I will now show the existence of the third Kähler structure in your case, when two Kähler forms have the same orientation, i.e. are sections of the bundle of, say, self-dual forms $\Lambda_{+}^2 TM$. I will first introduce a connection, and show that locally the third basis self-dual form is also Kahler. If $J_1, J_2$ are defined globally, then the global existence follows.
First of all, it is a standard fact that if $J_1$ is a Kähler form, then other self-dual 2-forms satisfy the following relations
$$
\begin{aligned}
\nabla_X J_2 &= P(X) J_3 \\
\nabla_X J_3 &= - P(X) J_2\, \qquad \forall X \in TM
\end{aligned}
$$
where $P$ is the potential of the Ricci form, i.e. $\mathcal{R}=d P$, see e.g., Derivative of ASD 2-forms on Kahler space, as well as
$$
\nabla_X J_1 = 0.
$$
If $J_2$ is another Kähler structure, then $P$ is closed, and Ricci form vanishes which implies that $\nabla_X J_3 = 0$ as well.
