With hindsight, this question is really too elementary for MO, in my opinion. However, since I am not on MSE, here is a sketch of what I think is a counter-example (but I think there is some merit to the argument that questions at this level should not be encouraged on MO). Also, I will not be happy if I learn later that this example is used in a paper without proper citation.
Note that the counter-example has nothing to do with ultraproducts, as I commented above you are trying to interchange limits and this hope is unfortunately far too naive.
$\newcommand{\norm}[1]{{\Vert#1\Vert}}$
Let $A=C[0,1]$ with usual multiplication and the supremum norm. Let $\phi\in\Delta_A$ be the character defined by $\phi(f)=f(0)$.
Let $(w_n) \subset A$ be a bounded sequence with the following properties: $w_n(0)=1=\norm{w_i}$ and $\operatorname{supp}(w_n)\subseteq [0,1/n]$ for all $n\geq 1$. (For instance $w_n(t) = \max(0,1-nt)$ would do.)
Claim 1: for any $a\in A$, $\lim_n\norm{aw_n - \phi(a)w_n} = 0$.
Now let $(a_k)\subset A$ be any bounded sequence with the following properties: for each $k\geq 1$, we have $a_k(0)=0$ and $a_k(t)=1=\norm{a_k}$ for all $t\in [1/k, 1]$. (For instance, $a_k(t) = \min(1, kt)$ would do.)
Claim 2: for each $n$, $\lim_k \norm{a_k w_n - \phi(a_k)w_n} = 1$.
Since $\lim_{k\in \mathcal U} \equiv \lim_k$ for any sequence of convergent complex numbers, this shows that the interchange of limits which you desire, is false even for a very well-behaved commutative Banach algebra. I leave the proofs of the two claims to you, the crucial point is of course that elements of $A$ are continuous functions on $[0,1]$.
On the other hand, note that every ultrapower of $C[0,1]$ is a commutative $C^*$-algebra, hence is amenable. So the result you actually want to prove, concerning $\phi_{\mathcal U}$-amenability of $A_{\mathcal U}$, may be true for some special cases of $A$ - it might even be true for all $A$, or it might not - but you have to come up with a better argument than the one you suggest in your question.
