I know you said you were looking for Plessner's original, but you can give a short proof of this. Fix a compactly supported smooth function $\phi$ with integral one. Let $\phi_{t}(x)=t^{-1}\phi(x/t),$ and set $\nu_{t}=\mu*\phi_{t}.$ Note that $\nu_{t}$ is absolutely continuous with respect to Lebesgue with Radon-Nikodym derivative $\int\phi_{t}(x-y)\,d\mu(y).$ If $f$ is a $C_{0}(\Bbb R)$ function a straight-foward computation using Fubini's Theorem shows that
$$\int fd\nu_{t}=\int\int f(y)\phi_{t}(y)d\tau_{x}\mu(y)dx$$
where $\tau_{x}\mu(E)=\mu(E-x).$ Because $\mu$ is continuous under translation the usual arguments show that $\nu_{t}$ converges to $\mu$ in norm, (here think of the total variation norm as the operator norm). Thus $\mu$ is a norm limit of $L^{1}(\Bbb R)$ functions and thus must be in $L^{1}(\Bbb R).$

(Note that the argument can be used to prove similar statements e.g. an $L^{\infty}$ function continuous under translations must agree almost everywhere with a uniformly continuous function).