Is the ideal gas constant not a constant if Celsius is used instead of Kelvin?
Charles' Law certainly DOES work with temperature in degrees Centigrade! 

If we express it (relative to $T=0$ absolute) as $V/T = k$ and if we convert $T$ (K) to $t$ (°C) as $t = T-273$ then obviously Charles' Law would have the form $V/(t+273) = k$.

More generally if, instead of $T$ we use some linear function of $T$, say $Y= mT + b$, then $$V/T = Vm/(Y-b) = k$$ or $$V/(Y-b) = k/m,$$ where $k$, $b$, and $m$ are constants. We can substitute $k'$ for $k/m$ and get $V/(Y-b) = k'$, but we can't get rid of the $-b$ that way.

The claim that a law (even if it is an approximate one like Charles') "doesn't work" with a different unit of measure confuses a formula with a law. Formulas are great for solving high school level problems, but a deeper understanding is expected in college and beyond. Any strictly increasing function can be used as a temperature scale, for instance $X  = π^T$ could be a temperature scale.

The formula we'd use to express Charles' Law would look very different, but the relationship between volume and $\log(X)/\log(π)$ would be just as constant (and since $\log(π)$ is a constant, we'd have $V/\log(X) = k''$.)