I'm answering Yemon's version of the question.

The answer is trivially yes for discrete $G$ since $\ell^1(G) \subset \ell^2(G)$, so let me focus on the non-discrete case.

The first observation to make is that $B(G)$ is contained in the bounded (and uniformly continuous) functions of $G$. So the question asks in particular if every integrable function on $G$ is the sum of a bounded function and a square-integrable function.

This is clearly false for compact infinite $G$: For such $G$ we have the strict inclusions $L^\infty \subsetneqq L^2 \subsetneqq L^1$ so $L^\infty + L^2 \subset L^2$, and hence every function in $L^1 \smallsetminus L^2$ provides a counterexample to the question.

Since the question asks for a counterexample in $\mathbb{R}^{n}$, I'll give one for $\mathbb{R}$ which is easily adapted to the higher-dimensional case and with a little care should also give a counterexample for any non-compact and non-discrete locally compact abelian group.

Take $f = \sum_{n=1}^{\infty} n \cdot [n,n+\frac{1}{n^{3}}]$. This is a function in $L^1 \smallsetminus L^2$. For a bounded function $h$ we have for all $n \geq \Vert h \Vert_{\infty}$ and all $x \in [n,n+\frac{1}{n^3}]$ that $|f(x) - h(x)| \geq n- \Vert h \Vert_{\infty}$, which implies that $g = f - h \notin L^2(\mathbb{R})$ by a straightforward estimate.