You can't have a convolution of two functions in $L_1(0,c)$ that is not in $L_1(0,c)$, else you violate Young's inequality, or the fact that $L_1(0,c)$ (i.e. the space you refer to as $L^{+}$) is a Banach algebra under convolution, so the answer is no. I assume you are taking $a < c$ and $b < c$ (otherwise you can't work in $L_1(\mathbb{R})$ because $f_a$ and $f_b$ don't belong there).

EDIT: Excuse me, Arc. I do apologize. I did not make the relevant distinction between an everywhere discontinuous function and a function defined only on a set of measure zero. I now understand the definition, so my answer is largely unhelpful.

However, if you are trying to construct a function that is discontinuous on a dense set, I do get the feeling that your construction wont do because the output of your convolution is (as you would expect) better behaved than the inputs, it is locally in $L_{2-\epsilon}$ and still has only one point of discontinuity. Perhaps I am missing the point here, though. Can you explain a bit more about what you want to achieve and your approach?