The answer is no (to the main question, not the title). Consider the *-algebra $S$ of *-polynomials generated by one variable $z$ such that $zz^* = z^*z$, *i.e.* the free commutative *-algebra on one generator. Each $a \in S$ can be considered to be a continuous function $\mathbb{C} \rightarrow \mathbb{C}$, so we can define
$$
\|a\|_1 = \sup \{ a(x) \mid x \in [0,1] \} \\
\|a\|_2 = \sup \{ a(x) \mid x \in [0,2] \}
$$
Since $[0,1] \subseteq [0,2]$, it is clear that $\|\cdot\|_1 \leq \|\cdot\|_2$. By the Stone-Weierstrass theorem, the completion of $S$ in $\|\cdot\|_1$ is $C([0,1])$ and the completion of $\|\cdot\|_2$ is $C([0,2])$.

So we can take a continuous function $b : [0,2] \rightarrow \mathbb{C}$ that vanishes on $[0,1]$ but takes a non-zero value in $(1,2]$, and a sequence $(a_i)_i$ in $S$ such that $a_i \to b$ with respect to $\|\cdot\|_2$. Then
$$
\|a_i\|_1 = \|a_i-b\|_1 \leq \|a_i - b\|_2 \to 0
$$
So $(a_i)_i$ is convergent with respect to both norms and has limit $0$ in $\|\cdot\|_1$, but its limit in $\|\cdot\|_2$ is $b \neq 0$.