No, it is not true unless the odd part is zero or the characteristic is 2. More precisely, there is no natural Hopf algebra structure outside these conditions, but there may be some odd examples. It is a Hopf superalgebra instead, which is what Peter May means in his answer.
You can see this on the easiest example: $L_0=0$, $L_1=F$, the field. Its universal enveloping is $F[x]/(x^2)$. It is a Hopf algebra if and only if the characteristic of $F$ is 2.
In general, you have difficulty with comultiplying the odd element $x\in L_1$ because $2x^2=[x,x]$. If you try $\Delta (y)=1\otimes y + y \otimes 1$ for $y\in L_0$ as suggested by Gods, no formula for $\Delta (x)$ will accommodate this relation.