$\def\one{\mathbb{1}}$ Here is a direct attack. First, we give a concise definition of the Frobenius character map, then we check that it is a map of rings.

Let $(\alpha_1, \alpha_2, \dotsc)$ be any sequence of nonnegative integers, all but finitely many of which are $0$. Define $\lvert\alpha\rvert = \sum \alpha_i$ and $x^{\alpha}=\prod x_i^{\alpha_i}$. For $\alpha$ with $\lvert\alpha\rvert =m$, define $S_{\alpha}$ to be the subgroup $\prod S_{\alpha_i}$ of $S_{m}$, embedded in the obvious way. (Note that there is no problem with the $\alpha_i=0$ terms; $S_0$ is the trivial group.)

For a representation $V$ of $S_m$, we define the Frobenius character of $S_m$ as
$$\sum_{\lvert\alpha\rvert=m} \dim V^{S_{\alpha}} x^{\alpha}.$$
(Here $V^{S_{\alpha}}$ is the $S_{\alpha}$-invariants in $V$.)
This is clearly additive. Also, permuting the elements of $\alpha$ conjugates the group $S_{\alpha}$, so this is a symmetric polynomial.

It remains to check multiplicativity.
In other words, let $V$ and $W$ be representations of $S_m$ and $S_n$ and let $\lvert\gamma\rvert = m+n$. Comparing the coefficient of $x^{\gamma}$ on both sides, we must check that
$$\dim \left(\operatorname{Ind}_{S_m \times S_n}^{S_{m+n}} V \boxtimes W \right)^{S_{\gamma}} = \sum_{\lvert\alpha\rvert = m,\ \lvert\beta\rvert = n,\ \alpha+\beta = \gamma} (\dim V^{S_{\alpha}}) (\dim W^{S_{\beta}}).\tag{$\ast$}\label{ast}$$

We write $\one$ for the trivial representation of any group. So, for $H \subset G$ and $V$ a representation of $G$, we have $\dim V^H = \dim \operatorname{Hom}_H(\one,\ \operatorname{Res}^G_H V)$. So the left hand side of \eqref{ast} is
$$\dim \operatorname{Hom}_{S_{\gamma}}\left(\one,\ \operatorname{Res}^{S_{m+n}}_{S_{\gamma}} \operatorname{Ind}_{S_m \times S_n}^{S_{m+n}} (V \boxtimes W) \right).$$

By Mackey's formula, the restriction–induction can be written as a sum over double cosets $S_{\gamma} \backslash S_{m+n} / (S_m \times S_n)$. In other words, we take permutation matrices of size $m+n$, break them into blocks where the row blocks have sizes given by $\gamma$ and column blocks have sizes $m$ and $n$, and quotient by permutation of rows and columns within blocks. It isn't hard to see that the double cosets are determined by how many ones are in each block. Letting $\alpha_i$ be the number of ones in the left block of the $i$-th row, and $\beta_i$ the number of ones in the right block of the $i$-th row, the double cosets are indexed by $(\alpha, \beta)$ with $\lvert\alpha\rvert = m$, $\lvert\beta\rvert = n$ and $\alpha+\beta = \gamma$.

The contribution of $(\alpha, \beta)$ is
$$\dim \operatorname{Hom}_{S_{\alpha} \times S_{\beta}} \left( \one, \operatorname{Res}_{S_{\alpha} \times S_{\beta}} (V \boxtimes W) \right) = \dim (V \boxtimes W)^{S_{\alpha} \times S_{\beta}}.$$
Of course one needs to check that all the confusing conjugacies in Mackey's formula simply say to let $S_{\alpha}$ and $S_{\beta}$ act in the obvious ways but, really, what else could they do?

We have $(V \boxtimes W)^{S_{\alpha} \times S_{\beta}} = V^{S_{\alpha}} \boxtimes W^{S_{\beta}}$ so the dimensions multiply. Thus, the $(\alpha, \beta)$ coset contributes exactly the $(\alpha, \beta)$ summand in \eqref{ast}, and we win. $\square$

No symmetric polynomial theory, no construction of Specht modules, just chasing induction and restriction!