= Solution
For each of the operators $D=d,\partial,\bar\partial,d^c$, define its nonnegative <Laplacian> by $\Delta_D=DD^*+D^*D$ using the $L^2$ <formal adjoint>. In particular, $\Delta=dd^*+d^*d$, $\Delta_\partial=\partial\partial^*+\partial^*\partial$, $\Delta_{\bar\partial}=\bar\partial\bar\partial^*+\bar\partial^*\bar\partial$, and $\Delta^c=d^c(d^c)^*+(d^c)^*d^c$. Use the <d c operator> convention
$$
d^c=i(\bar\partial-\partial),\qquad dd^c=2i\partial\bar\partial.
$$
This normalization, without an extra factor of one-half, is the one for which the requested equality with $\Delta^c$ holds.
The <Lefschetz operator of a Kähler manifold> is $L\alpha=\omega\wedge\alpha$, and the <adjoint Lefschetz operator> is its pointwise metric adjoint $\Lambda=L^*$. It lowers type by $(1,1)$ and is a real operator because $\omega$ and the metric are real. Write $P=\partial$ and $B=\bar\partial$, use $[A,D]=AD-DA$, and use $\{A,D\}=AD+DA$ for the anticommutators of the odd-degree operators. The allowed relation among the <Kähler identities> and its complex conjugate give
$$
[\Lambda,P]=iB^*,\qquad
[\Lambda,B]=-iP^*,\qquad
B^*=-i[\Lambda,P],\quad P^*=i[\Lambda,B].
$$
These identities alone suffice for the calculation. Since $P^2=B^2=0$,
$$
\{P,B^*\}=-i\{P,[\Lambda,P]\}
=-i(P\Lambda P-P^2\Lambda+\Lambda P^2-P\Lambda P)=0.
$$
Replacing $P$ by $B$ in the same calculation gives $\{B,P^*\}=0$. Thus both mixed terms vanish when expanding the <Hodge Laplacian> of $d=P+B$:
$$
\Delta=\Delta_P+\Delta_B.
$$
The anticommutation $PB=-BP$ now proves equality of the two summands. Indeed,
$$
\begin{aligned}
\Delta_B&=-i\{B,[\Lambda,P]\}
=-i(B\Lambda P+PB\Lambda+\Lambda PB-P\Lambda B),\\
\Delta_P&=i\{P,[\Lambda,B]\}
=i(P\Lambda B-PB\Lambda-\Lambda PB-B\Lambda P),
\end{aligned}
$$
which are the same expression. Finally, $(d^c)^*=i(P^*-B^*)$, so its <Laplacian> expands as
$$
\Delta^c=\Delta_B+\Delta_P-\{B,P^*\}-\{P,B^*\}=\Delta_B+\Delta_P.
$$
This proves the <Kähler Laplacian identity>, with its real-operator version included:
$$
\boxed{\Delta=2\Delta_{\bar\partial}=2\Delta_\partial=\Delta^c.}
$$
The action of the <almost complex structure> on forms is $J\alpha(v_1,\ldots,v_k)=\alpha(Jv_1,\ldots,Jv_k)$. It multiplies a form of type $(p,q)$ by $i^{p-q}$. Since $\Delta=2\Delta_{\bar\partial}$ preserves each bidegree, it commutes with this action of $J$. Thus $\Delta(J\alpha)=J(\Delta\alpha)$; the invertibility of $J$ gives
$$
\boxed{\Delta\alpha=0\quad\Longleftrightarrow\quad\Delta(J\alpha)=0.}
$$
This argument applies to arbitrary sums of types, not only to pure-type forms.
The <ddbar lemma> for a compact <Kähler manifold> states that a $d$-closed form of pure type is $\partial\bar\partial$-exact if it is $d$-exact, $\partial$-exact, or $\bar\partial$-exact. Conversely, a $\partial\bar\partial$-exact form is $d$-closed and exact for all three operators. In particular a $d$-exact, $d$-closed $(p,q)$-form equals $\partial\bar\partial\beta$ for a form $\beta$ of type $(p-1,q-1)$.
The difference $\gamma=\widetilde\omega-\omega$ of the two <Kähler forms> is a real $(1,1)$-form, is $d$-closed, and is $d$-exact by equality of their <de Rham cohomology> classes. The <ddbar lemma> therefore gives a smooth complex-valued function $h$ with $\gamma=\partial\bar\partial h$. Reality and anticommutation imply
$$
\gamma=\bar\gamma
=\bar\partial\partial\bar h=-\partial\bar\partial\bar h.
$$
Set $f=(h-\bar h)/(2i)$, which is real. It follows directly that $i\partial\bar\partial f=\gamma$, giving the <global potential for cohomologous Kähler forms>
$$
\boxed{\widetilde\omega=\omega+i\partial\bar\partial f,\qquad f\in C^\infty(X,\mathbb R).}
$$
If $f_1,f_2$ are two real potentials, $u=f_1-f_2$ satisfies $\partial\bar\partial u=0$, so it is a <pluriharmonic function>. The allowed relation among the <Kähler identities> gives $\bar\partial^*\bar\partial u=-i[\Lambda,\partial]\bar\partial u=-i\Lambda\partial\bar\partial u=0$, because $\Lambda\bar\partial u=0$ by degree. The <Kähler Laplacian identity> then gives $\Delta u=2\bar\partial^*\bar\partial u=0$. Integration by parts on compact $X$ gives
$$
0=\langle\Delta u,u\rangle_{L^2}=\|du\|_{L^2}^2,
$$
so $u$ is constant on each connected component. Consequently \b[the potential is unique up to one additive constant when $X$ is connected; on a disconnected $X$, one constant per component is the exact conclusion]. Without the usual connectedness convention the printed uniqueness assertion needs this qualification: distinct constants on the components of a disjoint union leave the same pair of <Kähler forms> unchanged.
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