It is not hard to show, that $(L^1(\mathbb R),*)$, where $(f*g)(t)=\int_{\mathbb R} f(t-s)g(s)ds$ has no unity element.
Consider therefore the functions $\phi_{\epsilon}=I_{[-\epsilon,\epsilon]}$. If $e$ is the unit, we get for almost all $t\in[-\epsilon,\epsilon]: 1=(e*\phi_{\epsilon})(t)=\int_{[-\epsilon,\epsilon]} e(t-s)ds$, which converges to zero for $\epsilon \rightarrow 0$.
My question is the following: $e$ seems to "want to be" a dirac delta but cannot. Is is possible to embed $L^1(\mathbb R)$ into a bigger space, which contains something similar to a dirac delta, which gives me a unit?
Surely. You can thin of elements of $L^{1}$ as (real or complex )measures and convolution of functions in $L^{1}$ is a special case of convolution of measures. The delta function is a unit element for measures under convolution. (To think of an $L^{1}$ function $f$ as mesure just consider $d\mu (x) =f(x)\,dx$).