Can you find a domain where ax+by=1 has a solution

Step 1
Define a domain ${\displaystyle R_0}$ as follows. Take a field K, adjoin an indeterminate ${\displaystyle x_0}$, and localize at ${\displaystyle \left(x_0\right)}$ (that is, adjoin inverses to everything not a multiple of ${\displaystyle x_0}$).
${\displaystyle R_0}$ has all its ideals principal and linearly ordered: ${\displaystyle \left(x_0\right)}$ contains ${\displaystyle \left(x_0^2\right)}$ contains ${\displaystyle \left(x_0^3\right)}$
Now given ${\displaystyle R_i}$ define ${\displaystyle R_{i+1}}$ inductively: Adjoin an indeterminate ${\displaystyle x_{i+1}}$, so we have ${\displaystyle R_i\left[x_{i+1}\right]}$. Quotient by ${\displaystyle (x_{i+1}^2-x_i)}$. Finally, localize at the prime ideal ${\displaystyle \left(x_{i+1}\right)}$
This effectively just gives us one more principal ideal containing all the principal ideals from ${\displaystyle R_i:\left(x_{i+1}\right)}$ contains ${\displaystyle (x_{i+1}^2=\left(x_i\right)}$ contains ${\displaystyle \left(x_i^2\right)}$
Now let R be the union of all the ${\displaystyle R_i}$, and its

The easiest example I know is the ring of all algebraic integers (roots of monic polynomials with integer coefficients).
As noted, it is a Bezout domain, so every finitely generated ideal is principal, and in particular for every two algebraic integers a and b there exist algebraic integers ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ such that ${\displaystyle \alpha a+\beta b=d}$, where d is a gcd for a and b. However, the ideal
${\displaystyle (2,2^{\frac{1}{2}},2^{\frac{1}{4}},2^{\frac{1}{8}},\cdots ,2^{\frac{1}{2^n}},\cdots )}$
is not principal, so the ring is not a PID.

Step 1
Given an integral domain R, we say that $a,b\in R$ are coprime if gcd(a,b) exists and $gcd(a,b)=1$.
On the other hand, we say that a and b are comaximal if there are $x,y\in R$ such that $ax+by=1$.
It's easy to see that comaximal $\Rightarrow$ coprime, but the other implication isn't necessarily true. Domains where coprime $\Rightarrow$ comaximal were called by Cohn Pre-Bézout domains.
As the names suggest, these aren't necessarily Bézout domains, because we only have the "Bézout relationship" for coprime elements.
But, it turns out that we can use Pre-Bézout domains to characterize Bézout domains among the class of GCD domains. More exactly, it's true the following.
Step 2
Theorem: Let R be an integral domain. TFAE:
i) R is a Bézout domain.
ii) R is a GCD Pre-Bézout domain.
Proof: i) $\Rightarrow$ ii) It's immediate.
ii) $\Rightarrow$ i)
Let $a,b\in R$
WLOG, we can suppose that $a\ne 0\ne b$
As R is a GCD domain, then $d=gcd(a,b)$ exists.
By an elementary property of gcds we have that $1=gcd(\frac{a}{d},\frac{b}{d})$ and since R is Pre-Bézout then $\frac{a}{d}$ y $\frac{b}{d}$ are comaximal, which means that there are $x,y\in R$ such that
$\frac{a}{d}x+\frac{b}{d}y=1$
Finally, if we multiply by d the above equality we get
ax+by=d
Thus d is a R-linear combination of a and b. Hence, R is Bézout domain.
In conclusion, according to Cohn, the class of domains you are looking for are known as Pre-Bézout domains, and these aren't necessarily Bézout domains, let alone PIDs.