No.

Set $M:=\mathcal B(\mathcal H)$ and let $(e_j)$ be an ONB for $\mathcal H$.
Set
$$ S_i(e_j) := \begin{cases} e_0, \quad &j=0; \\ -e_0,\quad &i=j; \\ 0, \quad &\text{otherwise} \end{cases} $$
and
$$ T_i(e_j) := \begin{cases} e_0+e_i, \quad &j=0; \\ 0, \quad &\text{otherwise} \end{cases} $$

Then $S_iT_i=0$.
The ultraweak limit of $(S_i)$ is the nonzero projection $S$ defined by
$$ S(e_j) := \begin{cases} e_0, \quad &j=0; \\ 0, \quad &\text{otherwise} \end{cases} $$
and $(T_i)$ has the same limit.

This answers your "more generally" question, but also your original question: Let $A$ be the universal C*-algebra generated by a 2 elements, $s,t$ of norm at most 2, satisfying $st=0$, then define $\phi_i(s):=S_i$ and $\phi_i(t):=T_i$. If $\phi$ is a point-ultraweak limit of the $\phi_i$ then $\phi(s)=S=\phi(t)$, so $\phi$ can't be multiplicative.