Bad proof that if $a + b + ab = 2020$ then $a+b=88$ Can you prove this:

Let $a,b \in \mathbb{N}$. If $a + b + ab = 2020$ then $a+b=88$.

This is the attempt given:
$\frac{2020-88}{a b}=1$
$a+b=88$
Substituting for $b$ using the $2$nd equation.
$2020-88 = a (88-a)$
That is a quadratic that is easily solved and gets $a = 42$ and $a = 46$.
So we have $a = 42$ and $b = 46$ or $a = 46$ and $b = 42$.
Was the question answered? A valid proof?

Key Idea $\ $ Completing a square generalizes to two-variables $\,x,y\,$ to completing a product
$$\begin{eqnarray} && xy + bx + cy\\
&=\,& x(y\!+\!b) + cy \\
&=\,& x(y\!+\!b) + c(y\!+\!b)-cb \\
&=\,& \qquad\, (x\!+\!c)(y\!+\!b) - cb
\end{eqnarray}\qquad\qquad\quad$$
${\rm So}\ \ b=1=c\ \ {\rm yields}\ \ n = xy\!+\!x\!+\!y = (x\!+\!1)(y\!+\!1)-1\!\iff\! n\!+\!1 = (x\!+\!1)(y\!+\!1)\ $ which has solutions $\,x,y = j\!-\!1,k\!-\!1\,$ for each factorization $\,n\!+\!1 = jk.\,$
Remark $\ $ In the same way we can show that if $\,a\ne 0\,$ then
$$\ axy\! +\!bx\!+\!cy = n\!\! \overset{\ \times\,a}\iff\!  (ax\!+\!c)(ay\!+\!b) = an\!+\!cb$$
Alternatively one can  mechanically derive the above from the first special case $(a = 1)$ using the $\rm AC$-method as described here.
