$\newcommand\R{\mathbb R}$$\newcommand\LBV{\mathrm{LBV}}$As was noted in comments, the set of all monotone functions (or your version of it,
$\mathrm{CM}^+(\R)$) is not a linear space.
An appropriate linear space here is $\LBV$, the space of all (say) right-continuous functions from $\R$ to $\R$ of locally bounded variation — because any function of locally bounded variation is the difference between two nondecreasing functions.
Take now any $f\in\LBV$. Let $\mu_f$ be the corresponding signed Lebesgue–Stieltjes measure, defined by the formula $\mu_f((a,b])\mathrel{:=}f(b)-f(a)$ for all real $a$ and $b$ such that $a<b$. It is easy to check that for all real $x$ we have
$$f(x)=f(0)+\int_{(0,\infty)}\mu_f(du)\,1(x\ge u)-\int_{(-\infty,0]}\mu_f(dv)\,1(x<v);$$
that is,
$$f=f(0)\,1+\int_{(0,\infty)}\mu_f(du)\,1_{[u,\infty)}-\int_{(-\infty,0]}\mu_f(dv)\,1_{(-\infty,v)}.$$
Thus, any $f\in\LBV$ is the limit of linear combinations of the following, say basic, functions: (i) the constant $1$, (ii) the functions of the form $1_{[u,\infty)}$ for real $u>0$, and (iii) the functions of the form $1_{(-\infty,v)}$ for real $v\le0$. In this sense, one might want to say that these basic functions form a basis of $\LBV$.
$\big($Because $1_{(-\infty,v)}=1-1_{[v,\infty)}$ for all real $v$, we have another, equivalent set of the basic functions of $\LBV$, consisting of the following functions: (i) the constant $1$ and (ii) the functions of the form $1_{[u,\infty)}$ for all real $u$.$\big)$
