The answer is positive under the assumptions boldfaced below, which hold at least when
$G$ is a complete connected Riemannian manifold (see the Remark at the end of this answer), so that the case of $G=\mathbb R^d$ is of course included.
On the other hand, the simple counterexample given by M. Dus shows that without a connectedness condition the desired conclusion will fail to hold in general.

Since the measure $\mu$ is nonzero, the set $E:=\{x\in G\colon f(x)>t\}$ is of nonzero $\lambda$-measure for some real $t>0$, where $f:=\frac{d\mu}{d\lambda}$. Also, $E=E^{-1}$, since $\mu$ is symmetric. Hence,
\begin{equation*}
f^{*2}(x)=\int f(y^{-1}x)f(y)dy\ge t\int_E f(y^{-1}x)\,dy\ge t^2|E\cap(xE^{-1})|=t^2|E\cap(xE)|,
\end{equation*}
where $dy:=\lambda(dy)$, and $|\cdot|:=\lambda(\cdot)$. By the regularity of the measure $\lambda$,
\begin{equation*}
|C_E|>\tfrac12\,|C|
\end{equation*}
for some compact set $C$, where $C_E:=C\cap E$.
**Assuming that $G$ has a countable basis of neighborhoods of $1_G$**,
it follows, by continuity, that
\begin{equation*}
|(xC)_E\cap C_E|>\tfrac12\,|C| \tag{1}
\end{equation*}
for all $x\in B$, where $B$ is
some (without loss of generality, symmetric) neighborhood of $1_G$. So,
\begin{equation}
f^{*2}(x)\ge t^2|(xE)\cap E|\ge t^2|(xC)_E\cap C_E|>\tfrac{t^2}2\,|C|=:c_1>0 \tag{2}
\end{equation}
for all $x\in B$.

**Suppose next that for some symmetric neighborhood $\hat B$ of $1_G$ such that $\hat B\subseteq B$ and all natural $n\ge2$ we have**
\begin{equation*}
a_n:=\inf_{x\in \hat B^n}\,|(xB)\cap \hat B^{n-1}|>0. \tag{3}
\end{equation*}
(If e.g. $G=\mathbb R^d$, $B$ is the ball centered at $0$ of "radius" $r>0$ with respect to (say) the $\ell_\infty^d$ norm, and $\hat B=\frac12\,B$, then $a_n=r^d$.)
Define recursively $c_n:=c_{n-1}c_1 a_n$ for $n\ge2$, with $c_1$ as above.

It follows by induction that
\begin{equation*}
f^{*2n}\ge c_n>0
\end{equation*}
on $\hat B^n$. Indeed, for $n=1$ this follows by (2) and the condition $\hat B\subseteq B$. For $n\ge2$ and any $x\in \hat B^n$,
by induction and (2),
\begin{multline*}
f^{*2n}(x)=\int f^{*2}(y^{-1}x)f^{*2(n-1)}(y)dy\ge c_{n-1}\int_{\hat B^{n-1}} f^{*2}(y^{-1}x)dy \\
\ge c_{n-1}\int_{(xB)\cap \hat B^{n-1}} f^{*2}(y^{-1}x)dy
\ge c_{n-1}c_1 a_n=c_n>0.
\end{multline*}
(For the penultimate inequality in the above display, we used the fact that for any $y\in xB$ we have $y^{-1}x\in B^{-1}=B$ and hence, by (2), $f^{*2}(y^{-1}x)\ge c_1$.)
Thus, the desired result follows if we **finally assume that for any compact $K$ and any (or, equivalently, any symmetric) neighborhood $U$ of $1_G$ there is a natural $n$ such that**
\begin{equation}
U^n\supseteq K. \tag{4}
\end{equation}

**Remark.** Conditions (1), (3), and (4), used above, will hold when $G$ is metrizable with a metric $d$ satisfying the following "linear" connectedness condition:
\begin{equation}
\forall x,y\in G\ \forall t\in[0,1]\ \exists z\in G\quad d(x,z)=(1-t)d(x,y)\quad \& \quad d(z,y)=td(x,y). \tag{5}
\end{equation}
In particular, condition (5) holds when $G$ is a complete connected Riemannian manifold; see e.g. wiki/Hopf--Rinow theorem and wiki/Riemannian_manifold.

Let us now verify that (5) implies (1), (3), and (4). This is obvious concerning (1), since $G$ is now a metric space and therefore has a countable basis of neighborhoods of $1_G$. Also, (5) implies
\begin{equation}
B_r^n=B_{nr} \tag{6}
\end{equation}
for all natural $n$ and all real $r>0$, where $B_r:=\{x\in G\colon d(x,1_G)\le r\}$, the $d$-ball of radius $r$ centered at $1_G$.

Take now any real $r>0$ and any compact $K$. Then $K\subseteq FB_r$ for some finite $F\subseteq G$. By (6), for each $x\in F$ there is some natural $n_x$ such that $x\in B_r^{n_x}$ and hence $xB_r\subseteq B_r^{n_x+1}$. So, $K\subseteq FB_r\subseteq B_r^n$ for $n:=1+\max_{x\in F}n_x$, which verifies (4).

It remains to verify (3). Without loss of generality, there we have $B=B_r$ for some real $r>0$. Take any natural $n\ge2$, then any $a\in(1/n,1)$, and let $\hat B:=B_{ar}$. It is enough to show that
\begin{equation}
a_n\ge|B_b|>0, \tag{7}
\end{equation}
where $b:=\frac r2\,(1-a)>0$.
First here, if $|B_b|=0$ then $|K|=0$ for any compact $K$, which would contradict the condition that $G$ is compactly generated. So, we have the second inequality in (7). To verify the first inequality there, take any $x\in\hat B^n=B_{nar}$, so that $d(x,1_G)=nar_1$ for some $r_1\in[0,r]$. Let
\begin{equation*}
s:=nar_1-\tfrac{r_1}2\,(1+a)=(n-1)ar_1-\tfrac{r_1}2\,(1-a)\in[0,(n-1)ar_1].
\end{equation*}
By (5), there is some $z\in G$ such that
\begin{equation*}
d(z,1_G)=s,\quad d(x,z)=nar_1-s=\tfrac{r_1}2\,(1+a).
\end{equation*}
For any $y\in zB_b$, we have $d(y,z)\le b$ and hence
\begin{equation*}
d(y,x)\le d(y,z)+d(z,x)\le b+\tfrac{r_1}2\,(1+a)\le b+\tfrac r2\,(1+a)=r,
\end{equation*}
\begin{equation*}
d(y,1_G)\le d(y,z)+d(z,1_G)\le b+s\le b+(n-1)ar-\tfrac r2\,(1-a)=(n-1)ar,
\end{equation*}
so that $y\in (xB_r)\cap B_{(n-1)ar}=(xB)\cap \hat B^{n-1}$.
So, $zB_b\subseteq(xB)\cap \hat B^{n-1}$, whence $|(xB)\cap \hat B^{n-1}|\ge|B_b|$, for all $x\in\hat B^n$,
which implies the first inequality in (7).

Thus, all the conditions (1), (3), and (4) are verified.