Background:
For any commutative ring $R$, let $\mathbf{Symm}_R$ be the ring of symmetric functions in countably many variables $x_1$, $x_2$, $x_3$, ... over $R$. ("Symmetric functions" really means symmetric power series of bounded degree.) It is known that $\mathbf{Symm}_R$ is generated by the elementary symmetric polynomials $e_1$, $e_2$, $e_3$, ... as an $R$-algebra. If $R$ is a $\mathbb Q$-algebra, then $\mathbf{Symm}_R$ is also generated by the power sum polynomials $p_1$, $p_2$, $p_3$, ... as an $R$-algebra. Note that $\mathbf{Symm}_{\mathbb Z} \subseteq \mathbf{Symm}_{\mathbb Q}$.
There are two ways to define a comultiplication (in the sense of coalgebras) on $R$:
The first comultiplication, called $\Delta_1$, is defined by $\Delta_1\left(e_n\right) = \sum\limits_{i+j=n} e_i\otimes e_j$ for all $n\in\mathbb N$, where the sum allows $i$ and $j$ to be zero (and $e_0$ has to be understood as $1$). This comultiplication satisfies $\Delta_1\left(f\left(x_1,x_2,x_3,...\right)\right) = f\left(x_1,x_2,x_3,...,y_1,y_2,y_3,...\right)$, where we identify the tensor product $\mathbf{Symm}_R\otimes_R\mathbf{Symm}_R$ as a ring of certain power series in "$2$ times countably many" indeterminates $x_1$, $x_2$, $x_3$, ..., $y_1$, $y_2$, $y_3$, .... In order to make sense of the term $f\left(x_1,x_2,x_3,...,y_1,y_2,y_3,...\right)$, one has to recall that $f$ is a symmetric polynomial, so that one can reorder its arguments in any way, for example as $x_1,y_1,x_2,y_2,x_3,y_3,...$.
The second comultiplication, denoted by $\Delta_2$, satisfies $\Delta_2\left(p_n\right) = p_n\otimes p_n$ for all positive integers $n$. This is not enough to define it because $p_1$, $p_2$, $p_3$, ... don't always generate the $R$-algebra $\mathbf{Symm}_R$, but at least they generate it when $R$ is a $\mathbb Q$-algebra, so one can use this definition for $R=\mathbb Q$, then show that $\Delta_2\left(\mathbf{Symm}_{\mathbb Z}\right) \subseteq \mathbf{Symm}_{\mathbb Z}\otimes_{\mathbb Z}\mathbf{Symm}_{\mathbb Z}$, so that $\Delta_2$ is also defined for $R=\mathbb Z$, and consequently (since $\mathbb Z$ is the initial object in the category of rings) also defined for any $R$. Of course, one could just as well give a more direct definition of $\Delta_2$, by setting
$\Delta_2\left(f\left(x_1,x_2,x_3,...\right)\right) = f\left(x_1y_1,x_1y_2,x_1y_3,...,x_2y_1,x_2y_2,x_2y_3,...,x_3y_1,x_3y_2,x_3y_3,...\right)$.
Again, one has to invoke symmetry of $f$ for the right hand side to make sense here. This time one also needs to check that the right hand side is well-defined at all, what with the infinitely many terms involving the same $x_i$. Yet another way to define $\Delta_2$ is by the identity $\Delta_2\left(e_n\right) = \sum\limits_{\lambda\text{ is a partition of }n} s_{\lambda}\otimes s_{\lambda^t}$, where $s_{\mu}$ are the Schur polynomials, and $\lambda^t$ is the conjugate partition of $\lambda$. (A categorification of this identity is the isomorphism from MathOverflow question #120873 posted earlier today.)
This suggests defining a third comultiplication $\Delta_3$ by
$\Delta_3\left(f\left(x_1,x_2,x_3,...\right)\right) = f\left(x_1+y_1,x_1+y_2,x_1+y_3,...,x_2+y_1,x_2+y_2,x_2+y_3,...,x_3+y_1,x_3+y_2,x_3+y_3,...\right)$.
This, however, doesn't go well: Setting $f=p_n=x_1^n+x_2^n+x_3^n+...$, the right hand side becomes $\sum\limits_{i\geq 1,\ j\geq 1} \left(x_i+y_j\right)^n$, which involves summing infinitely many $x_i^n$'s for every $i\geq 1$, and summing infinitely many $y_i^n$'s for every $i\geq 1$.
Fortunately, these are the only undefined terms on the right hand side; all the other infinite sums do make sense. If we replace every undefined sum of infinitely many $x_i^n$'s by $rx_i^n$ for some fixed integer $r$, then we obtain
$\Delta_3\left(p_n\right) = \sum\limits_{i=1}^{n-1} \dbinom{n}{i} p_i \otimes p_{n-i} + r \otimes p_n + p_n \otimes r$.
Question:
So fix an integer $r$, and let us define a map $\Delta_3 : \mathbf{Symm}_{\mathbb Q} \to \mathbf{Symm}_{\mathbb Q} \otimes_{\mathbb Q} \mathbf{Symm}_{\mathbb Q}$ by
$\Delta_3\left(p_n\right) = \sum\limits_{i=1}^{n-1} \dbinom{n}{i} p_i \otimes p_{n-i} + r \otimes p_n + p_n \otimes r$ for every positive integer $n$.
This $\Delta_3$ is easily seen to be coassociative, and for $r=1$ even counital (with respect to the standard counity of $\mathbf{Symm}_{\mathbb Q}$).
Do we have $\Delta_3\left(\mathbf{Symm}_{\mathbb Z}\right) \subseteq \mathbf{Symm}_{\mathbb Z}\otimes_{\mathbb Z}\mathbf{Symm}_{\mathbb Z}$ for every $r$ ? In other words, can this $\Delta_3$ be defined over any commutative ring $R$ ? What is the combinatorial meaning of this $\Delta_3$ ?
Evidence:
We need to prove that $\Delta_3\left(e_n\right)$ has integer coefficients for all $n$ and $r$. Sage code and some output (please don't imitate my code) verifies my suspicion for $r=0,1$ and $n=1,2,3,...,9$. I also have some not very tangible semi-proof arguments.
