If we replace $P$ by its closure, it has the same boundary and the projection onto $L$ has the same inner points. Thus we may assume without loss of generality that $P$ is closed.
Let $L_p$ be the line projected onto a point $p$ of the projection. Since $P$ is convex and closed, the set of points on $L_p$ that belongs to $P$ is a closed interval. We want to show that this interval is not a single point if $p$ is an inner point of the projection.
If the interval were a single point for all points $p$ in the projection, then $P$ would have to be a line, which contradicts the assumption that $\dim\operatorname{aff}P=2$. Thus there is at least one point $q$ in the projection for which the interval contains two distinct points. For any inner point $p$, we can find a point $r$ in the projection on the other side of $p$ than $q$. Since $r$ is in the projection, at least one point $a$ in $P$ is projected onto $r$. Since $P$ is convex, connecting the two distinct points in the interval for $q$ with $a$ yields two distinct points in $P$ that are projected onto $p$. Thus the interval on $L_p$ belonging to $P$ cannot be a single point.