This might help: If $u$ and $v$ are both close enough to scalars in the operator norm,
and $\varepsilon$ is small enough, then no matter how large $k$ is, if the group generated by $u^{\prime}$ and $v^{\prime}$ is finite, then it still has to be Abelian. The reasoning is something like this: I use the operator norm : given a unitary matrix $M$, there is a unique scalar $\lambda$ ( in the interior of the unit disk unless $M$ itself is a scalar matrix) with $\|M- \lambda I\|$ minimal, and this only depends on the spectrum of $M$. In particular, we can only have $\|M -\lambda I\| < \frac{1}{2}$ for a scalar $\lambda$ if the eigenvalues of $M$ all lie on an arc of length less than $\frac{\pi}{3}$ on the unit circle.It is essentially a Theorem of Frobenius that if $G$ is a finite group of unitary matrices, then the subgroup $H$ generated by the matrices at distance less than $\frac{1}{2}$ from a scalar is an Abelian normal subgroup. If $u$ and $v$ are close enough to scalars, and $\varepsilon$ is small enough then we can make both $\|u^{\prime} - \lambda I\|$
and $\| v^{\prime} - \mu I\|$ less than $\frac{1}{2}$ for suitable scalars $\lambda,\mu$. So if they
generate a finite group, it must be Abelian. Note added: I will add a little more detail.
One proof of Jordan's theorem uses a compactness argument and a kind of contraction Lemma.
Let $[a,b]$ denote $a^{-1}b^{-1}ab$. If $a$ and $b$ are unitary, then
$$\| I-[a,b] \| = \|I - a^{-1}b^{-1}ab \| = \|ab-ba \|$$
$$ = \| (a-I)(b-I) - (b-I)(a-I)\|
\leq 2 \|a-I\| \|b-I\|$$. Hence if $\| I-a\| < \frac{1}{2}$, then $\|I- [a,b] \| < \|I-b\|$
for all $b.$ Hence in a finite group $G$ of unitary matrices, let $H$ be the subgroup generated
by those elements of $G$ with $\|I-h\|< \frac{1}{2}$. For any $h \in H$ with
$\|I - h \| < \frac{1}{2}$, we have $[g,h,h, \ldots ,h] = 1$ for all $g \in G$. By a theorem of Baer, $H$ is nilpotent. With a little more work, you actually get $H$ Abelian. (This is a kind of hybrid of arguments of Frobenius and a little more finite group theory. More or less the same line of reasoning led to a similar theorem by
Zassenhaus on discrete groups, and this led to the idea of a Zassenhaus neighbourhood in a
Lie group). Note that if $x,y$ are in different cosets of $H$ in $G$, we must have
$\|x-y\| \geq \frac{1}{2}$, so the compactness of the unit ball of $M_{n}(\mathbb{C})$ gives
a fixed bound on $[G:H]$. But the same argument works when $\| a - \lambda I \| < \frac{1}{2}$
for some scalar $\lambda$, using $$\|ab - ba \| = \| (a-\lambda I)(b-\mu I) - (b-\mu I)(a-\lambda I)\|$$ for any scalars $\lambda,\mu$, and replacing $H$ by the group generated by elements at
distance less than $\frac{1}{2}$ from any scalar. (Added in response to comment below): You do need a little argument to get from $H$ nilpotent to $H$ Abelian, but the statements above are all accurate. It is important to remember that $H$ is generated by elements within distance $\frac{1}{2}$ of a scalar. Any such element has all its eigenvalues on an arc of length less than $\frac{\pi}{3}$ on $S^{1}$. No subset of the "multiset" of eigenvalues of such an element has zero sum (it's clear eg, if you rotate so that all eigenvalues have positive real part). Let $\chi$ be the character of the given representation of $H$. Let $\mu$ be an irreducible constituent of $\chi$. By the remark above, if $\|h -\alpha I \| < \frac{1}{2}$ for some scalar $\alpha$, then $\mu(h) \neq 0.$ But if $\mu$ is not linear, it is induced from a character of a maximal subgroup $M$, which is normal as $H$ is nilpotent. But then $\mu$ vanishes on all elements of $H$ which are outside $M$. Hence the elements at distance less than $\frac{1}{2}$ from a scalar must all lie within $M$. But since $M$ is proper, and such elements generate $H$, this is a contradiction. Hence $\mu$ must be linear after all. Since $\mu$ was arbitrary, $H$ is Abelian.