$\newcommand{\epi}{\operatorname{epi}}$What you are trying to prove is not true without additional assumptions on $f$ and $g$.
Allow me to denote the infimal convolution of $f$ with $g$ by $f\# g$ because it renders better in MathJax.
Firstly, we can show that:
Proposition 1. $\epi f + \epi g \subseteq \epi(f \# g)$.
Proof. If we take $(x, a) \in \epi f$ and $(x', a') \in \epi g$, then
\begin{align}
(f \# g)(x+x') = \inf_{x+x' = z_1 + z_2} f(z_1) + g(z_2) \leq f(x) + g(x') \leq a + a'.
\end{align}
which comples the proof. $\Box$
Proposition 2. If the infimal convolution is exact, that is, if we can write $(f \# g)(x) = \min_{y} f(y) + g(x-y)$, then $\epi (f\# g) = \epi f + \epi g$.
Proof. Take $(x, c) \in \epi (f\# g)$. It suffices to show that $(x, c) \in \epi f + \epi g$. If the infimal convolution $f\# g$ is exact, then for given $x$, there exists a $y^* = y^*(x)$ so that
\begin{align}
(f \# g)(x) = f(y^*) + g(x-y^*),
\end{align}
therefore,
\begin{align}
{}&{}f(y^*) + g(x-y^*) \leq c
\\
\Leftrightarrow{}&{}g(x-y^*) \leq c - f(y^*)
\\
\Leftrightarrow{}&{} \left(x-y^*, c - f(y^*) \right) \in \epi g
\end{align}
Then,
\begin{align}
(x,c) = (y^*, f(y^*)) + \left(x-y^*, c - f(y^*) \right) \in \epi f + \epi g,
\end{align}
which completes the proof. $\Box$
Proposition 3. If $f\# g$ is exact, then $(f\# g)(x) = \inf_{\lambda \in \mathbb{R}}\{(x, \lambda) \in \epi f + \epi g\}$.
Proof. In general, for any function $F:\mathbb{R}^n \to \mathbb{R}$, it is
$$F(x) = \inf_{\lambda \in \mathbb{R}} \{F(x) \leq \lambda\} = \inf_{\lambda \in \mathbb{R}} \{(x, \lambda) \in \epi F\}.$$
This is known as epigraphical relaxation. For the infimal convolution we have
\begin{align}
(f\# g)(x) {}={}& \inf_{\lambda \in \mathbb{R}} \{(x, \lambda) \in \epi (f\# g)\}
\\
{}={}& \inf_{\lambda \in \mathbb{R}} \{(x, \lambda) \in \epi f + \epi g\}.
\end{align}
Note: There are some conditions under which the infimal convolution is exact, for example if $f$ and $g$ are lsc and convex and $f$ satisfies $\lim_{t\to\infty} f(x)/\|x\| = \infty$, then $f\# g$ is exact.
Best Answer
I finally found a solution to the problem
$$ g*(y) = \underset{x}{\sup}\; \{x^Ty - g(x)\} $$
As we know
$$ g(y) = \underset{x_1 + x_2 = x}{\inf}\; \{f_1(x_1) + f_2(x_2)\} $$
Now we have
$$ g*(y) = \underset{x}{\sup}\; \{x^Ty - \underset{x_1 + x_2 = x}{\inf}\; \{f_1(x_1) + f_2(x_2)\} \} $$
$$ = \underset{x}{\sup}\; \{x^Ty + \underset{x_1 + x_2 = x}{\sup}\; \{-f_1(x_1) - f_2(x_2)\} \} $$
$$ = \underset{x}{\sup}\underset{x_1 + x_2 = x}{\sup}\; \{x^Ty -f_1(x_1) - f_2(x_2)\} \} $$
$$ = \underset{x_1,x_2}{\sup}\; \{x_1^Ty + x_2^Ty -f_1(x_1) - f_2(x_2)\} \} $$
$$ = f_1^\star(x_1) + f_2^\star(x_2) $$