= Solution
Let $G$ be locally compact, Abelian and Hausdorff, and fix a <Haar measure>. Write $T_xf(y)=f(x^{-1}y)$ and use <convolution>
$$
(f*g)(y)=\int_G f(x)g(x^{-1}y)\,dx.
$$
The <L1 group algebra> is a commutative <Banach algebra>. Here a <group character> means a continuous <homomorphism> $\chi:G\to\mathbb T$, not the trace of an arbitrary group representation. A <Gelfand character> is a nonzero multiplicative complex-linear functional on the <convolution> algebra. We use the convention
$$
\boxed{\Phi_\chi(f)=\int_G f(x)\chi(x)\,dx}.
$$
With the alternative standard Fourier convention involving $\overline\chi$, one simply conjugates every <group character> in this identification.
For such $\chi$, the functional is bounded by $\|f\|_1$, and <Fubini's theorem>, the substitution $y=xz$, and the <homomorphism> law give
$$
\Phi_\chi(f*g)=\int\!\int f(x)g(z)\chi(xz)\,dz\,dx
=\Phi_\chi(f)\Phi_\chi(g).
$$
It is nonzero: if $u\ge0$ is nonzero in $C_c(G)$, then $f=\overline\chi u$ has $\Phi_\chi(f)=\int u>0$.
Conversely let $\Phi$ be an <algebra character>. It is automatically bounded, with $|\Phi(f)|\le\|f\|_1$. Indeed, extend it to the <unitization> by $\widetilde\Phi(f+\lambda1)=\Phi(f)+\lambda$. If $|\Phi(f)|>\|f\|_1$, the <Neumann series> makes $f-\Phi(f)1$ invertible, whereas its character value is zero, contradicting multiplicativity of an invertible element and its inverse.
Choose $g$ with $\Phi(g)\ne0$ and put $\chi(x)=\Phi(T_xg)/\Phi(g)$. Commutativity of the group gives $(T_xf)*g=f*(T_xg)$; applying $\Phi$ yields
$$
\Phi(T_xf)=\chi(x)\Phi(f)\quad\text{for every }f.
$$
Thus the ratio is independent of the chosen $g$. Applying the relation successively to $T_xT_yg=T_{xy}g$ gives $\chi(xy)=\chi(x)\chi(y)$, and $\chi(e)=1$. In particular it never vanishes.
The translation map $x\mapsto T_xg$ is continuous in $L^1$. For $g\in C_c(G)$, translated supports near a fixed $x$ lie in one <compact set>; continuity on that set and a finite cover give <uniform convergence> of the translated values, hence $L^1$ convergence because the <compact set> has finite <Haar measure>. For general $g$, approximate by $C_c(G)$ and use $\|T_xg\|_1=\|g\|_1$. Therefore $\chi$ is continuous. For all integers $n$,
$$
|\chi(x)|^n=\frac{|\Phi(T_{x^n}g)|}{|\Phi(g)|}
\le\frac{\|g\|_1}{|\Phi(g)|}.
$$
Positive powers exclude modulus greater than one; negative powers exclude modulus less than one. Hence $|\chi(x)|=1$.
Finally, the $L^1$-valued <convolution> formula $f*g=\int f(x)T_xg\,dx$ and boundedness of $\Phi$ give
$$
\Phi(f)\Phi(g)=\Phi(f*g)=\int f(x)\Phi(T_xg)\,dx
=\Phi(g)\int f(x)\chi(x)\,dx.
$$
Cancel $\Phi(g)$ to obtain the displayed formula for every $f$. Different characters give different functionals: if their continuous difference is nonzero at a point, integration against a suitably phased compactly supported test function in a small <neighbourhood> detects that difference. Thus the correspondence is bijective.
It also identifies the <character space of an Abelian L1 group algebra> topologically with the <Pontryagin dual group>. For a net converging uniformly on <compact subsets> in $\widehat G$, convergence of the functionals follows first for compactly supported $f$, then for all $L^1$ functions using the common <norm> bound one. Conversely suppose the functionals converge in the <Gelfand topology>. Choose $g$ whose limiting character value is nonzero; its values are eventually bounded away from zero. For compact $K\subset G$, the set $\{T_xg:x\in K\}$ is compact in $L^1$. Convergence of uniformly bounded <linear functionals> is uniform on this <compact set>: a finite norm-net reduces it to convergence at finitely many functions, with the common bound controlling the approximation error. The ratio formula then gives <uniform convergence> of the <group characters> on $K$. This proves equality with the <compact-open topology>, without assuming first countability of $G$.
For later use, these transforms form a self-adjoint subalgebra dense in $C_0(\widehat G)$. The set of <algebra characters> together with the zero functional is weak-star compact by <Banach-Alaoglu theorem> and the closed multiplicativity equations. Removing zero gives a locally compact Hausdorff <character space>; each evaluation $\Phi\mapsto\Phi(f)$ vanishes at infinity, since its modulus-at-least-$\epsilon$ set is compact and avoids zero. Transforms separate characters, and none vanishes identically at a given character. They are closed under <conjugation> because $f^*(x)=\overline{f(x^{-1})}$ satisfies $\Phi_\chi(f^*)=\overline{\Phi_\chi(f)}$. The locally compact <Stone-Weierstrass theorem> therefore gives the claimed density.
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