$a/p+b/q>1$ with certain conditions

Your concerns are not unfounded.

First of all, if $0 < a < p$ and $0 < b < q$, which certainly is possible, then $$\frac{a}{p} + \frac{b}{q} < \frac{p}{p} + \frac{q}{q} = 2\,.$$ It follows that in your argument, if everything works up to that point, you cannot have both of $x$ and $y$ be positive integers. More precisely it follows that $x + y \leqslant 1$.

We can explicitly parametrise the pairs $(m,n)$ such that $pn + aq = qm + bp$: One immediately verifies that $m_0 = a$, $n_0 = b$ works. Since changing $m$ changes the right hand side by a multiple of $q$ and changing $n$ changes the left hand side by a multiple of $p$, any two possible choices of $m$ differ by some multiple of $p$ and any two possible choices of $n$ differ by some multiple of $q$. Hence we have $m = m_0 + sp$ and $n = n_0 + tq$ for some integers $s,t$. Plugging this in, we see that we have a solution if and only if $s = t$, thus the solutions to $pn + aq = qm + bp$ are the pairs $(m_t,n_t)$, $t \in \mathbb{Z}$, where $m_t = a + tp$ and $n_t = b + tq$.

Now writing $$\frac{a}{p} + \frac{b}{q} = \frac{(a - m_s) + m_s}{p} + \frac{(b - n_t) + n_t}{q} \tag{1}$$ is certainly possible, we don't even need to have $s = t$ in that. But to have $$\frac{(a - m_s) + m_s}{p} + \frac{(b - n_t) + n_t}{q} \geqslant \frac{a - m_s}{p} + \frac{b - n_t}{q} \tag{2}$$ we need $$\frac{m_s}{p} + \frac{n_t}{q} \geqslant 0 \tag{3}$$ which only obviously holds if $s \geqslant 0$ and $t \geqslant 0$. However, in that case the right hand side of $(2)$, namely $(-s) + (-t)$, is nonpositive, thus certainly not $> 1$. A less obvious but still easy to see condition for $(3)$ is $s + t \geqslant 0$, but that again renders the right hand side of $(2)$ nonpositive. We are left with the option $s + t = -1$ (see above, $s = -x$, $t = -y$), but then $(3)$ is equivalent to $$\frac{a}{p} + \frac{b}{q} \geqslant 1\,.$$ Since $p$ and $q$ are distinct primes, this inequality is necessarily strict (if it holds), thus we are back at square one — to make the argument work, we must first prove what the argument is supposed to prove.

So let's take a different route. Get rid of the denominators, i.e. multiply the inequality with $pq$. Thus we want to show $$aq + bp > pq\,. \tag{$\ast$}$$ Now we know everything in $(\ast)$ is a positive integer, which can be useful. And we know some congruences: \begin{align} aq + bp &\equiv aq \equiv 1 \pmod{p}\,,\\ aq + bp &\equiv bp \equiv 1 \pmod{q}\,. \end{align} Since $p$ and $q$ are coprime, it follows that $$aq + bp \equiv 1 \pmod{pq}\,. \tag{$\ast\ast$}$$ And also $aq + bp \geqslant 1\cdot q + 1\cdot p > 1$. Together with $(\ast\ast)$ it follows that $aq + bp \geqslant pq + 1$, which is $(\ast)$.

(And in case $a < p$, $b < q$ we have $aq + bp \leqslant (p-1)q + (q-1)p < 2pq$, hence $aq + bp = pq+1$.)