Your conjecture is correct, and here is why.
A key property of $\delta$ is that it preserves the total degree (recall that the total degree of a monomial $x_0^{e_0}x_1^{e_1}\ldots x_n^{e_n}$ is $\sum_{k=0}^n e_k$). If we denote by $R_t$ (for $t\geq 0$) the subspace of $R$ generated by all the monomials of total degree exactly $t$, then $R$ is the direct sum of the $R_t$ and each $R_t$ is stable by $\delta$, so the study of $\delta$ on $R$ reduces to the study of $\delta$ on each $R_t$. From now on, we fix a $t\geq 0$ and restrict ourselves to $R_t$.
Let $\cal S$ be the subpace generated by $S \cap R_t$. Our goal is to show that any monomial $m\in R_t$ can be uniquely written $m=s+d$ with $s\in {\cal S}$, $d\in \delta R_t$.
Uniqueness: That's the easy part, we must show that ${\cal S} \cap \delta R_t =\lbrace 0 \rbrace$. Let $d\in \delta R_t$ with $d\neq 0$. Then $d=\delta r$ for some nonzero $r\in R_t$. Let $n$ be the largest integer such that $x_n$ appears in $r$, i.e. the degree $d_n$ of $x_n$ in $r$ is positive. Then $d=\delta r$ contains one or several monomials of the form (monomial in $x_0,x_1,\ldots,x_{n-1}$)$x_n^{d_n-1}x_{n+1}$ ; those monomials are not in $S$, so $d\not\in {\cal S}$.
Existence: For this part, as noted in a comment in the OP, we must be careful in the choice of a monomial ordering, as the obvious, easy choices don't work.
Let $m$ be a monomial in $R_t$. We can write $m=x_{i_1}x_{i_2}\ldots x_{i_t}$ with $i_1\leq i_2\leq \ldots \leq i_t$. For $2\leq k\leq t$, put $d_k=i_k-i_1$. Then all the $d_k$'s are nonnegative. Next, define
$$
\mu(m)=(d_2+d_3+\ldots+d_t,d_t,d_{t-1},\ldots,d_2,i_1) \tag{1}
$$
By construction, this $\mu(m)$ is in ${\mathbb N}^t$. Denote by ${\leq}_1$ the lexicographic (dictionary) order in ${\mathbb N}^t$, and let ${\leq}_2$ denote the well-ordering of monomials of $R_t$ define by $m {\leq}_2 m'$ iff $\mu(m) {\leq}_1 \mu(m')$.
We then show that each monomial $m$ is in ${\cal S}+\delta R_t$, by ${\leq}_2$-induction on $m$, or in other words by ${\leq}_1$-induction on $\mu(m)$.
Base case : the ${\leq}_1$ ordering in ${\mathbb N}^t$ starts with the initial segment $\lbrace (0,0,\ldots,0,i) \rbrace$ for $i\in {\mathbb N}$, corresponding to the monomials $m_i=x_i^t$ for $i\in {\mathbb N}$. When $t=1$ and $i>0$, this $m_i$ monomial can be decomposed as $m_i=0+\delta(x_{i-1})$, otherwise it is in $S$ and can therefore be decomposed as $m_i=m_i+\delta(0)$.
Induction step : suppose that there is a decomposition for all $m'$ such that $\mu(m') {\leq}_1 \mu(m)$. Write $m=x_{i_1}x_{i_2}\ldots x_{i_t}$ with $i_1\leq i_2\leq \ldots \leq i_t$. If $i_{t-1}=i_t$, then $m\in S$ and we are done. So we can assume $i_{t-1} < i_t$. Let $M=x_{i_1}x_{i_2}\ldots x_{i_{t-1}}x_{i_t-1}$. Let $d$ be the degree of $x_{i_t-1}$ in $M$, so that
$$
M=x_{i_1}x_{i_2} \ldots x_{i_{t-d}} (x_{i_t-1})^d \tag{2}
$$
It follows that
$$\delta(M)=\sum_{k=1}^{t-d} l_k +dm \tag{3}$$
where $l_k$ is the monomial defined by
$$
l_k = x_{i_1}\ldots x_{i_{k-1}}x_{i_k+1}x_{i_{k+1}}\ldots x_{i_{t}} \tag{4}
$$
From (3), we deduce $m=\delta(\frac{M}{d})-\frac{1}{d}\sum_{k=1}^{t-d}l_k$, so it will suffice to show that each $l_k$ is in ${\cal S}+\delta R_t$. By the induction hypothesis, it will suffice to show that $\mu(l_k) \lt_1 \mu(m)$ for each $k$.
Let $s$ be the largest integer $s\geq k$ such that $i_k=i_{k+1}=\ldots =i_{s}$. Then $s \lt t$, and we can write $l_k=x_{I_1}x_{I_2}\ldots x_{I_t}$, with $I_1\leq I_2\leq \ldots I_t$, with
$$
I_j=\left\lbrace\begin{array}{lcl}
i_j, & \textrm{when} & j\not\in \lbrace s,t \rbrace, \\
i_j+1, & \textrm{when} & j = s, \\
i_j-1, & \textrm{when} & j = t.
\end{array}\right. \tag{5}
$$
Suppose first that $s>1$. Then $\mu(l_k)=(D_2+D_3+\ldots+D_t,D_t,D_{t-1},\ldots,D_2,I_1)$ where
$$
D_j=\left\lbrace\begin{array}{lcl}
d_j, & \textrm{when} & j\not\in \lbrace s,t \rbrace, \\
d_j+1, & \textrm{when} & j = s, \\
d_j-1, & \textrm{when} & j = t.
\end{array}\right. \tag{6}
$$
In particular, the smallest index at which $\mu(l_k)$ and $\mu(m)$ differ is the second, and the corresponding coordinate in $\mu(l_k)$ is one less than in $\mu(m)$. So $\mu(l_k) \lt_1 \mu(m)$ as wished.
We are now left with the case $s=1$. In that case, $\mu(l_k)=(D_2+D_3+\ldots+D_t,D_t,D_{t-1},\ldots,D_2,I_1)$ where
$$
D_j=\left\lbrace\begin{array}{lcl}
d_j-1, & \textrm{when} & j\not\in \lbrace 1,t \rbrace, \\
d_j+1, & \textrm{when} & j = 1, \\
d_j-2, & \textrm{when} & j = t.
\end{array}\right. \tag{7}
$$
In particular, the smallest index at which $\mu(l_k)$ and $\mu(m)$ differ is the first, and the corresponding coordinate in $\mu(l_k)$ is $t$ less than in $\mu(m)$. So $\mu(l_k) \lt_1 \mu(m)$ again, which finishes the proof.
