Yes, the inequality is true. Indeed, the inequality in question can be rewritten (or, if you prefer, generalized) as
\begin{equation}
Eg(Y)\ge\ln(1-a^3), \tag{10}\label{10}
\end{equation}
where
$$g(t):=\ln(1+t)+(1-t)\ln(1-t)$$
and $Y$ is a random variable (r.v.) such that
\begin{equation}
P(|Y|\le a)=1\quad\text{and}\quad EY=0. \tag{20}\label{20}
\end{equation}
(One can take $Y:=f(X)$, where $X$ is a r.v. uniformly distributed on $[0,1]$.)
By well-known results (see e.g. this paper or this paper or Corollary 13 or formula (2.13)), without loss of generality (wlog) the distribution of the r.v. $Y$ is supported on a set $\{u,v\}\subset[-a,a]$ of cardinality $\le2$. Given such $u$ and $v$, if $a$ is now replaced by the smallest possible value of $b$ such that $\{u,v\}\subseteq[-b,b]$, the left-hand side of \eqref{10} will not change whereas the right-hand side of \eqref{10} will not decrease. So, wlog one of the values $u$ or $v$ coincides with one of the endpoints of the interval $[-a,a]$; that is, wlog either $v=-a$ or $v=a$. To cover both of these cases at once, assume wlog that the distribution of the r.v. $Y$ is supported on a set of the form $\{-a,u,a\}$ with $u\in[-a,a]$.
So then, it remains to show that
\begin{equation*}
L:=pg(-a)+qg(u)+rg(a)-\ln(1-a^3)\ge0 \tag{30}\label{30}
\end{equation*}
given the conditions
\begin{equation*}
-a\le u\le a,\quad p,q,r\ge0,\quad \\
p+q+r=1,\quad p(-a)+qu+sa=0. \tag{40}\label{40}
\end{equation*}
Solving the latter two equations in \eqref{40} for $p$ and $r$ and substituting the resulting expressions for $p$ and $r$ into \eqref{30}, we get
\begin{equation*}
L=M(q,a,u):=P(a)+qR(a,u),
\end{equation*}
where
\begin{equation*}
\begin{aligned}
P(a)&:=a \tanh ^{-1}(a)+\ln \left(1-a^2\right)-\ln \left(1-a^3\right) \\
Q(a,u)&:=g(u)-(1-u/2) \ln \left(1-a^2\right)-a \tanh ^{-1}(a).
\end{aligned}
\end{equation*}
Note that $M(q,a,u)$ is affine in $q$.
So, to prove \eqref{30} given \eqref{40}, it suffices to show that
\begin{equation*}
M(0,a,u)\ge0\quad\text{and}\quad M(1,a,u)\ge0 \tag{50}\label{50}
\end{equation*}
for $u\in[-a,a]$ and $a\in(0,1)$.
We have $M(0,a,u)=P(a)$,
$P(0)=0=P'(0)$ and
\begin{equation*}
P''(a)=H(a):=\dfrac{a \left(6+10 a+2 a^2-3 a^3+2 a^4+a^5\right)}{(1-a)^2 (1+a)^2
\left(1+a+a^2\right)^2}, \tag{60}\label{60}
\end{equation*}
which is obviously $>0$ for $a\in(0,1)$. So, $M(0,a,u)=P(a)>0$ for $a\in(0,1)$.
Next,
\begin{equation*}
M(1,a,u)=F(a):=F(a,u):=g(u)-\ln \left(1-a^3\right) \\
+\frac u2\, \ln \left(1-a^2\right),
\end{equation*}
\begin{equation*}
F'(a)\frac{(1 - a) (1 + a) (1 + a + a^2)}a
=3 a (1 + a) - (1+ a +a^2) u \\
\ge3 a (1 + a) - (1+ a +a^2) a
=a (2 + 2 a - a^2)>0.
\end{equation*}
So, $F$ is increasing in $a$ and hence $M(1,a,u)=F(a,u)\ge F(|u|,u)$, because $|u|\le a$. If now $u\in[0,1)$, then $F(|u|,u)=F(u,u)=P(u)\ge0$, by what was shown in the previous paragraph.
Finally, if $u\in(-1,0)$, then $F(|u|,u)=F(-u,u)=h(u):=g(u)+\frac u2\, \ln(1 - u^2) - \ln(1 + u^3)$, $h(0)=0=h'(0)$, $h''(u)=H(|u|)>0$ -- cf. \eqref{60}; so, $M(1,a,u)=F(a,u)\ge F(|u|,u)=F(-u,u)=h(u)\ge0$.
Thus, \eqref{50} is proved. $\quad\Box$
