This is nice, and I do not know a reference for this.

The notion of $p$-concavity is mentioned, for example, in Section 9 of

*Gardner, R. J.*, **The Brunn-Minkowski inequality**, Bull. Am. Math. Soc., New Ser. 39, No. 3, 355-405 (2002). ZBL1019.26008.

Note that a natural interpretation of $0$-concavity is log-concavity. Your result generalizes the fact that the product of log-concave functions is log-concave.

Also your result is equivalent to the following.

**Theorem.** *If functions $F$ and $G$ are concave, then so is $F^tG^{1-t}$ for all $t \in [0,1]$.*

Indeed, substitute $f^\alpha = F$, $g^\beta = G$, $t = \frac{\beta}{\alpha+\beta}$.

The concavity of $F^tG^{1-t}$ can be proved as follows. For every $\lambda \in (0,1)$ and $\mu = 1-\lambda$ the concavity of $F$ and $G$ imply
\begin{multline*}
F^tG^{1-t}(\lambda x + \mu y) = F^t(\lambda x + \mu y) G^{1-t}(\lambda x + \mu y)\\
\ge (\lambda F(x) + \mu F(y))^t (\lambda G(x) + \mu G(y))^{1-t}\\
\ge (\lambda F(x))^t(\lambda G(x))^{1-t} + (\mu F(y))^t(\mu G(y))^{1-t}\\
= \lambda F^tG^{1-t}(x) + \mu F^tG^{1-t}(y)
\end{multline*}
(In the middle we have used the inequality $(1+u)^t(1+v)^{1-t} \ge 1+u^t v^{1-t}$, for which I have only a nasty proof.)

One can prove the theorem also by computing the second derivative of $F^tG^{1-t}$ (which is fun), but the above argument is more general because it does not require differentiability.

I think these results should be known, but have no reference.

EDIT: As Alexei Kulikov pointed out, one can prove the Theorem by first proving the special case $t=\frac12$ as a lemma (in this case the nasty inequality becomes $\sqrt{(1+u)(1+v)} \ge 1 + \sqrt{uv}$, which follows from Cauchy-Schwarz). From the special case one infers by induction all $t = p/2^n$ cases, which implies the general case by a limit argument.

EDIT: A reformulation of the theorem: the space of exp-concave functions is convex.