The MO thread that t.b. links to doesn't seem to quite answer the original question, which if I've understood it correctly is:
do there exist $f,g\in L^1({\mathbb R})$, neither of which is a.e. zero, such that $f*g$ is a.e. zero?
I am pretty sure that the answer is yes, because the answer to the analogous question for $\ell^1({\mathbb Z})$ is yes. To see this, work on the circle and consider the simplest function $h_1$ which is piecewise linear on $[-\pi,\pi]$, is zero on the intervals $[-\pi,-\delta]$ and $[\delta,\pi]$ for some fixed $\delta$, and which is $1$ at $0$. (Triangular spike.) Then viewed as a continuous function on ${\mathbb R}/2\pi{\mathbb Z}={\mathbb T}$, the Fourier series of $h_1$ is absolutely summable. Now let $h_2(x) = h_1(x+\pi)$, clearly $h_1\cdot h_2=0$, which means that (FS of $h_1$) $*$ (FS of $h_2$) $=0$.
Now your question takes place on the line, not the circle, but my feeling is that the same construction should work. That is, let $h_1$ and $h_2$ be the same functions as before, but regard them as functions on ${\mathbb R}$. Then both should be the Fourier transforms of integrable functions $f_1,f_2\in L^1{\mathbb R})$, and once again $f_1*f_2=0$ since $h_1\cdot h_2=0$.
(Actually, thinking about it a bit more, I "know" that the answer to your question must be yes, because it is equivalent to the question "does the Fourier algebra $A({\mathbb R})$ contain non-trivial zero divisors?", and it's known that $A({\mathbb R})$ is a regular Banach algebra of functions, meaning that one can find functions which are zero on any prescribed closed subset $F$ and 1 on any prescribed compact set disjoint from $F$; pointwise product in $A({\mathbb R})$ corresponds to convolution in $L^1({\mathbb R})$.)