$(x,y)\subset \mathbb Q[x,y]$ is not a principal ideal

Its pretty much there, but I think a little more is needed, for example:

"The only event in which this happens is when $f(x,y)=ax, g(x,y)=1/a$"

we could have the other way around also right? For example $f(x,y) = 1$, $g(x,y) = x$. You should say also why this can't happen!

I second Alex J Best's comment about a little proof fix, but also I think it's worth noting that this holds in $R[x,y]$ for $R$ any commutative ring with identity. So $\mathbb{Q}$ is not special in this regard. Your proof will continue to work for any integral domain. But if $R$ has zero divisors, you cannot immediately say that $f$ has degree $1$, and the proof is a little more complicated.

You might like to try to prove the following for a general ring $R$

$f \in R[x,y]$ divides $x$ and $y$ iff $f$ is a unit

This immediately implies that $(x,y)$ is not principal. I include a proof sketch in the spoiler below if you are interested. Like your proof, it mostly involves comparing coefficients.

Proof: Write $fg = x$ and $fh = y$. Let $f_x, f_y$ and $f_0$ denotethe $x$ coefficient, $y$ coefficient, and constant coefficient of $f$. First we can show that $f_0$ is a unit. Suppose not the case. Looking at $1 = f_0g_x + f_xg_0$ and $0 = f_0 g_y + f_y g_0$ modulo $(f_0)$, we deduce $f_x \notin (f_0), f_y \in (f_0)$. But similarly we could do the same with $fh = y$ to deduce $f_y \notin (f_0), f_x \in (f_0)$. Together these yield a contradiction, so conclude $f_0$ is a unit.

Now, knowing that $f_0$ is a unit, compare coefficients in $fg = x$ to see that $g \in R[x]$, and similarly $h \in R[y]$. Considering $fg = 0 \text{ mod } x$ we see that $x$ divides $g$, and similarly $y$ divides $h$, which gives us the equation $yg = xh$. Comparing coefficients in $yg = xh$ we finally see $g = x$ and $h = y$ up to unit multiples. It follows that $f$ is a unit.

Your proof is almost fine, it just needs a couple of details. But you're overcomplicating things.

Consider $f(x,y)g(x,y)=x$ as an equality between polynomials in $y$ over $\mathbb{Q}[x]$. As this has degree $0$, both $f$ and $g$ must have degree $0$. Thus $b=0$. Similarly, $a=0$.

Therefore $f(x,y)=c$, which either generates the whole ring or the zero ideal. On the other hand, $(x,y)$ is a nonzero proper ideal, as $\mathbb{Q}[x,y]/(x,y)\cong\mathbb{Q}$.