No: your argument would apply equally well to any family of distributions, not just the family of distributions with finite mean & variance, & it's easy to come up with counterexamples where the sample variance is not a component of the sufficient statistic (e.g. the family of gamma distributions having various scales & shapes, for which the sample arithmetic & geometric means are jointly sufficient). Sufficient statistics of fixed dimension are updateable (see When if ever is a median statistic a sufficient statistic? for why the sample median can never be sufficient) but the converse doesn't follow.
With i.i.d. samples from the non-parametric family you specify, the order statistic $(X_{(1)}, \ldots, X_{(n)})$ is minimal sufficient—only the order of the observations lacks information about the distribution from which they arise. It's also complete: consequently, the sample mean and variance, while not sufficient themselves, as functions of the order statistic are not only unbiased estimators of their population analogues, but the unique uniformly minimum-variance unbiased estimators.
If you know $(\bar X_n, S^2_n)$ is sufficient for a sample of size $n$, then $(\bar X_{n+1}, S^2_{n+1})$ is sufficient for a sample of size $n+1$. If you can show the latter statistic is a function of $(\bar X_n, S^2_n, X_{n+1})$, which is trivial, it follows that $(\bar X_n, S^2_n, X_{n+1})$ is also sufficient, as @whuber points out.