Not always.

Let $M = \mathbb{R}^2 \setminus \mathbb{R}^+$ with the induced metric (plane with a slit). Let $U = M$ and $\varphi$ be the polar coordinate map $\varphi = (r,\theta)$. Let $r = x^1$ and $\theta = x^2$.

Your condition 1 requires that the level sets of $x^2$ in the primed coordinate system be the same as the level sets of $\theta$. But this means regardless of what you choose as the coordinate function $x^1$, you must have $\|\partial_2'\| \to \infty$ as "$r\to \infty$", due to the infinite separation of leaves of $\theta$ near infinity.

Observing that defining the distribution of $\{ \partial_1, \ldots, \partial_{n-1}\}$ is in fact equivalent to defining the level sets of the coordinate $x^n$ (modulo topological concerns), you can rephrase your question as the following:

Give a Riemannian manifold $U$ with metric $g$, and a non-degenerate function $v:U\to\mathbb{R}$ (in the sense that $\mathrm{d}v \neq 0$), can you find a vector field $\eta$ on $U$ such that $g(\eta,\eta) = 1$ and $\eta(v) = 1$.

The objection above the fold comes from the minmax characterisation:

$$ \|\mathrm{d}v\| = \max_{\eta: \|\eta\| = 1} \eta(v) $$

so that if $\|\mathrm{d}v\| < 1$ the problem cannot be solved; now, the $v$ should be a function with the same level sets as $x^n$, so if $\|\mathrm{d}x^n\|$ is not bounded below (from zero) initially, there is no hope of rescaling $v$ to satisfy the characterization.

Now, what if $\mathrm{d}v$ is bounded from below? by rescaling we can assume that it is bounded from below by 1. Next, since in the original question the level sets of $v$ are coordinated, we can assume that there exists also a non-vanishing vector field $\tau$ such that $\tau(v) = 0$ on our manifold $U$. Since $\tau$ is non-vanishing we can further assume that $g(\tau,\tau) = 1$.

Then to construct the vector field $\eta$ we can take the ansatz

$$ \eta = \frac{1}{\|\mathrm{d}v\|^2}\mathrm{d}v^\sharp + \alpha \tau $$

But construction $\eta(v) = 1$. And we just need to solve for $\alpha$ using the *algebraic* equation

$$ \|\eta\| = 1 \iff \frac{1}{\|\mathrm{d}v\|^2} + \alpha^2 = 1 \iff \alpha = \pm \sqrt{ 1 - \frac{1}{\|\mathrm{d}v\|^2}} $$