$\newcommand{\R}{\mathbb R}\newcommand{\EE}{\mathcal E}\newcommand{\de}{\delta}\newcommand\ep\varepsilon $The answer is yes, and the key here is the Dunford--Pettis theorem (Theorem 3.1; for more original and complete sources, see Dunford--Pettis, Theorem 3.2.1 and Bogachev, Theorem 4.7.20). According to this theorem, the condition that the sequence $(f_n)$ converges weakly in $L^1(\R)$ implies that this sequence is equi-bounded and equi-integrable. So, (i)
\begin{equation*}
M:=\sup_n\|f_n\|_1<\infty \tag{1}\label{1}
\end{equation*}
and (ii) for each real $\ep>0$ there exist a measurable set $A_\ep\subseteq\R$ with Lebesgue measure $|A_\ep|<\infty$ and a real number $\de_\ep>0$ such that
\begin{equation*}
\sup_n\int_{A_\ep^c}|f_n|\le\ep \tag{2}\label{2}
\end{equation*}
and
\begin{equation*}
\sup_n\sup_{E\in\EE_\de}\int_E |f_n|\le\ep, \tag{3}\label{3}
\end{equation*}
where $A_\ep^c:=\R\setminus A_\ep$ and $\EE_\de$ is the set of all measurable subsets $E$ of $\R$ with Lebesgue measure $|E|\le\de$.
Without loss of generality (wlog), the weak limit $f$ of the sequence $(f_n)$ is $0$, so that
\begin{equation*}
I_n(h):=\int_\R f_n h\to0
\end{equation*}
for all $h\in L^\infty(\R)$ (as $n\to\infty$). We have to show that
\begin{equation*}
J_n(h):=\int_\R g_n h\to0
\end{equation*}
for all $h\in L^\infty(\R)$.
For any $h\in L^\infty(\R)$, we have
$$J_n(h)=\int_\R g_n(x)h(x)\,dx \\
=\int_\R f_n(x-a_n)h(x)\,dx=\int_\R f_n(y)h(y+a_n)\,dy$$
and hence
\begin{equation*}
J_n(h)-I_n(h)=D_n(\R,h), \quad\text{where}\quad D_n(A,h):=\int_A f_n(x)[h(x+a_n)-h(x)]\,dx.
\end{equation*}
It remains to show that $D_n(\R,h)\to0$ for all $h\in L^\infty(\R)$.
Take any real $\ep>0$, and let $A_\ep$ and $\de_\ep$ be as in \eqref{2} and \eqref{3}. By \eqref{2},
\begin{equation*}
|D_n(\R,h)-D_n(A_\ep,h)|\le2\ep\|h\|_\infty. \tag{4}\label{4}
\end{equation*}
So, it is enough to show that $D_n(A_\ep,h)\to0$ for all $h\in L^\infty(\R)$.
The measurable set $A_\ep$ of finite measure can be approximated in measure by a finite union of finite intervals. So, by \eqref{3}, it is enough to show that
\begin{equation*}
D_n(l,h)\overset{\text{(?)}}\to0 \tag{5}\label{5}
\end{equation*}
for any finite interval $l$ and any $h\in L^\infty(\R)$. Let $L$ be a finite interval containing $l\cup\bigcup_n(a_n+l)$.
The function $h\in L^\infty(\R)$ can be uniformly approximated on the interval $L$ by a linear combination of indicators of measurable sets. So, in view of \eqref{1}, wlog $h=1_B$ for some measurable subset $B$ of the interval $L$.
The measurable set $B$ can be approximated in measure by a finite union of finite intervals. So, again by \eqref{3}, wlog $B=[u,v]$ for some real $u$ and $v$, and then $h=1_{[u,v]}$.
Therefore and, again, in view of \eqref{3},
\begin{equation*}
|D_n(l,h)|
\le\int_{[u,v]+[u-a_n,v-a_n]} |f_n|\to0,
\end{equation*}
where $[u,v]+[u-a_n,v-a_n]$ denotes the symmetric difference of $[u,v]$ and $[u-a_n,v-a_n]$.
Thus, \eqref{5} follows, as desired.
Clearly, the proof above will work for $L^1(\R^d)$ instead of $L^1(\R)$, by using Cartesian products of intervals instead of intervals.
