Note that if you already know the uniqueness criterion for CRT then the sought result is immediate, i.e. both $\,\color{#0a0}{x_0 = 1}\,$ and the CRT formula $\,x_1 = p(p^{-1}\!\bmod q) + q(q^{-1}\!\bmod p)$ are clearly solutions of the system $\,x\equiv 1\pmod{\!p},\ x\equiv 1\pmod{\!q},\,$ so by $\rm\color{#c00}{
U =}$ uniqueness $\,x_1\overset{\rm\color{#c00}U}\equiv \color{#0a0}{x_0\equiv 1}\pmod{\!pq}$.
Your question essentially asks for a (direct) proof of the uniqueness criterion. It is easy to show that the uniqueness criteria for [C]CRT, linear diophantine equations, and modular inverses are all equivalent (to Euclid's lemma and many other closely related basic results). Below is a direct proof via inverse uniqueness (with a trivial one-line proof) & $\rm\color{#0a0}{Bezout\ identity}$ for $\gcd(p,q)=1$.
There are $\,p',q'\in\Bbb Z\,$ with $\,\color{#0a0}{pp'\!+\!qq' = 1},\,$ so $\, \color{0a0}{pp'\equiv 1}\pmod{\!\color{darkorange1}q},\,$ so by inverse uniqueness $\,p^{-1}\equiv p'\!\pmod{\!q},\,$ so $\, p^{-1^{\phantom{|^|}}}\!\!\! = p'\!+\!iq,\,$ some $\,i\in\Bbb Z.\,$ Similarly $\,q^{-1^{\phantom{|^|}}}\!\!\! = q'\!+\!jp,\,$ some $\,j\in \Bbb Z,\,$ therefore
$\, \color{#90f}{pp^{-1}\!+qq^{-1^{\phantom{|^|}}}}\!\!\! = p(p'\!+\!iq)+q(q'\!+\!jp)$ $ = pp'\!+\!qq'^{\phantom{|^|}}\!\!\!+\color{#c00}{pq}(i\!+\!j)\:\!\equiv_{\color{#c00}{pq}}\:\!\color{#0a0}{pp'\!+\!qq'\, [= 1]}$
Remark $ $ No answer actually "deals with the $\color{#0af}{1\!+\!1}$" -- which arise from your change from working with inverses $p^{-1},q^{-1}$ to working with their $\color{#c00}{{\rm negatives}\ \,k,k'}\,$ (which has the following effect on the Bezout identity summands: $\,i+j = 1\iff \color{#c00}{-i}+(\color{#c00}{-j})+\color{#0af}{2}=1).\,$ Rather, we eliminate the more unwieldly negative inverses by changing back to $\,p^{-1},q^{-1},\,$ i.e. read your last displayed equations in reverse
to get $\,\color{#c00}kp+\color{#c00}{k'}q+\color{#0af}2 = \color{#90f}{p^{-1}p+q^{-1}p},\,$ which is $\color{#0a0}{\equiv 1\pmod{pq}}\,$ as proved above.
As usual, these matters are clarified arithmetically when viewed via the (product) ring form of CRT (see here for a simple intro requiring no knowledge of ring theory). Namely, your change to use of $\rm\color{#c00}{negated}$ inverses $\,\color{#c00}{k\equiv_q -p^{-1}}\,$ and $\,\color{#c00}{k'\equiv_p -q^{-1}}\,$ corresponds to a basis change to the negated standard basis $\,(0,\color{#c00}{-1}),(\color{#c00}{-1},0)\,$ vs. the normal standard basis $\,(0,1),(1,0).\,$ Then your final equation arises by negating the Bezout equation $\rm E$ then adding $\,\color{#0af}2,\,$ i.e. $\,\rm E\to -E\to \color{#0af}2+E,\,$ i.e.
$$\begin{align}
\ \overbrace{(0,1)\ +\ (1,0)\ \equiv \ (1,1)}^{\textstyle\!\! \color{#90f}{p^{-1}p\ +\,\ q^{-1}q}\ \equiv\ 1\ \ \ \ \ }
\!\overset{\rm -E}\iff \ \overbrace{(0,\color{#c00}{-1})+ (\color{#c00}{-1},0)\ \equiv \ (\color{#c00}{-1,-1})}^{\textstyle\!\!\!\!\!\!\!\!\!\!\!\!\!\!\color{#c00}{-p^{-1}}p\, \ \color{#c00}{-\ \ q^{-1}}q\, \ \ \equiv \ \ {-}1} \\[.7em]
\underset{\rm \color{#0af}2\,+\,E^{\phantom |}\!}\iff\ \ \underbrace{\color{#0af}{(2,2)}+ (0,\color{#c00}{-1})+(\color{#c00}{-1},0)\ \equiv\ (1,1)}_{\textstyle\!\!\! \color{#0af}{2}\ \ \ +\ \ \ \color{#c00}kp\ \ \ \ +\ \ \ \ \color{#c00}{k'}q\ \ \equiv\ \ 1}\quad\ \
\end{align}\qquad\qquad$$
Thus the three above congruences $\!\bmod{pq}\,$ are all trivially equivalent, so it suffices to prove any one of them. It is easiest to prove the first (as we did above) since it connects more immediately to the intuitive arithmetical notion of inverses and their uniqueness. As is often true in algebraic proofs, a slight algebraic transformation of equations may transform them into a form which better reveals innate arithmetical structure facilitating intuition and proof. As such we should strive to transform towards such intuitive forms - not away from them (as do the above transforms to negated inverses and basis).
