It seems to me that the set of projective semi-distances that you describe in the example (let's call them of type I) can be slightly generalized this way: Take as $H$ in the definition of $d_H$ any open half-plane, plus an open half-line contained in its boundary. In other words, up to an affinity, these other distances (say of type II) are defined by $d(x,y)=0$ or $1$ according whether $x \mathbf{\ge} y$ or not, where $\ge$ denotes the lexicographic order in $\mathbb{R}^2$ (we may also parametrize all of them by the unit tangent bundle of $\mathbb{R}^2$). In contrast, the distance of type I are analogously related to the order $x_1\ge y_1$. These two classes cover all $\{0,1\}$-valued projective semi-distances, and are all indecomposable (even in a slightly sense).
Indeed, given a semi-distance $d$ we may consider the equivalence relation $x\sim y\ \mathrm{iff}\ d(x,y)=0$, whose classes are the closures of points. (Just for chatting, let me remark that if $d$ is a semi-distance that is topologically stronger than a semi-distance $d'$, then the partition into closures of points induced by $d $ refines the partition induced by $d'$). If $d$ is projective, the closure of any point is a convex set (a convex combination of a pair of points at zero distance, must be also at zero distance from them).
If $d$ is a $\{0,1\}$-valued projective semi-distance, then the set of points at distance $1$ from any given point $p$ is also a convex set: for, if $d(p,x)=d(p,y)=1$ and $z:=tx+(1-t)y$ for some $0\le t\le 1$, then since $d(x,z)+d(z,y)=d(x,y)\le1$, at least one among $d(x,z)$ and $d(y,z)$ is zero, so that in any case $d(p,z)=1$.
But a convex set of $\mathbb{R^2}$ whose complement is also convex is necessarily either an open half-plane, or an closed half plane, or an open hals-plane plus an (either open or closed) half-line contained in its boundary. As a consequence, the partition induced by a $\{0,1\}$-valued projective semi-distance is necessarily into two classes, and therefore is of the above types I or II.
A $\{0,1\}$-valued projective semi-distance $d$ that is topologically stronger than a non-zero semi-distance $d'$ refines the partition of the $d'$-closures of points. But this easily implies that $d'$ induces the same partition, and that it can only take two values. Thus it is a positive multiple of $d$. In particular, if $d=d_1+d_2$, both $d_1$ and $d_2$ are multiples of $d$, (even if we do not assume they are projective).