**One construction.**

$G$ is a locally compact Hausdorff topological group.

Let $C_c(G)$ denote the collection of continouous, real-valued functions on $G$ with compact support.

(1) We begin with a *Haar integral*, a linear
functional $\Lambda : C_c(G) \to \mathbb R$. The Haar
integral is unique up to a constant factor.

(2) Then we construct
a set-function $\mu_1$ on the open subsets of $G$: If $U \subseteq G$
is open, let
$$
\mu_1(U) = \sup\{\Lambda(f) : f \in C_c(G), 0 \le f \le 1_U\} .
$$
Here, $1_U$ is the indicator function (characteristic function) of the set $U$.

(3) Now we construct a set function $\mu_2$ on all sets.
If $E \subseteq G$, let
$$
\mu_2(E) = \inf\{\mu_1(U) : U \text{ open, } U \supseteq E\} .
$$
This $\mu_2$ is a Carathéodory outer measure.

(4) This version of the *Haar measure* is the restriction $\mu_3$
of $\mu_2$ to the collection $\mathcal M$ of $\mu_2$-measurable sets.

So, I understand the question to be: If $E \in \mathcal M$, then is there
a Borel set $B$ and a $\mu_3$-null set $N$ so that $E = B \Delta N$?

**This Example**

Now consider the case $G = \mathbb R \times \mathbb R_d$.
The Haar integral we will use is:
$$
\Lambda(f) = \sum_{y \in \mathbb R_d} \int_{\mathbb R} f(x,y)\,dx .
$$
If $f \in C_c(G)$, then for all but finitely many $y$
the integral is identically zero (so it is a finite sum),
and for the remaining $y$, the integrand has compact
support in $\mathbb R$.

Next: an open set $U \subseteq G$ is obtained by arbitrarily choosing
open sets $U_y \subseteq \mathbb R$, one for each $y$, and taking
$$
U = \bigcup_y \big(U_y \times \{y\}\big) .
$$
For such $U$ we get
$$
\mu_1(U) = \sum_y \lambda(U_y) .
$$
Here, $\lambda$ is Lebesgue measure in $\mathbb R$. Note that
an uncountable sum has value $\infty$ unless all but countably
many terms are zero. And the only open set in $\mathbb R$
with measure zero is the empty set. So $\mu_1(U) < \infty$
implies that $U_y = \varnothing$ except for countably many $y$.

Now compute $\mu_2$. An arbitrary set $E \subseteq G$ is of
course obtained by taking arbitrary sets $E_y \subseteq \mathbb R$,
one for each $y$, and then
$$
E = \bigcup_y \big(E_y \times \{y\}\big) .
$$
Let $U \supseteq E$ open, so that
$U_y \supseteq E_y$ open for all $y$.
If $E_y = \varnothing$ for all but countably many $y$, we get
$$
\mu_2(E) = \sum_y \lambda(E_y)
$$
by taking the $U_y$ close to the corresponding $E_y$.
But if $E_y \ne \varnothing$ for uncountably many $y$, we get
$\mu_2(E) = \infty$, even if the series $\sum_y \lambda(E_y)$
has finite value. So, for example, if $E$ is an uncountable
subset of the $y$-axis in $G$, then $\mu_2(E) = \infty$
even though $\lambda(E_y)=0$ for all $y$. (This is used
for a standard counterexample to show a limitation in
Fubini's theorem.)

What about $\mathcal M$? Write $\mathcal L$ for the collection
of Lebesgue measurable sets in $\mathbb R$.
Let $E$ be an arbitrary set in $G$ as before.
We have $E \in \mathcal M$ if and only if $E_y \in \mathcal L$
for all $y$. *** proof omitted ***

Write $\mathcal B$
for the collection of Borel sets in $\mathbb R$. If $E$ is Borel in $G$
then $E_y \in \mathcal B$ for all $y$. (The converse is false, but
we won't need it.)

Now, consider a set $Q \subseteq [0,1]$ in
$\mathcal L \setminus \mathcal B$. The example to consider is
$E = Q \times \mathbb R_d$. So that $E_y = Q$ for all $y$.
Can it be that $E = B \Delta N$ with $B$ Borel and $N$ null?
Since $Q \not\in \mathcal B$, we would need
$N_y \ne \varnothing$ for all $y$. But then
$\mu_2(N) = \infty$ and it is not a null set after all.

**Conclusion** ... for this particular construction of Haar measure,
the question has a negative answer.