This problem was proposed by Burii.

Assume the hypothesis is false, and that, $ca-c+b\mid cb-c+a$. Then, $ca-c+b\mid cab-ca+a^2$. Moreover, we also have $ca-c+b\mid cab-cb+b^2$. This implies, $ca-c+b\mid ca+b^2-cb-a^2$. Observe also that, $ca+b^2-cb-a^2=(b-a)(b+a)-c(b-a) =(b-a)(b+a-c)>0$. In particular, this yields that, $$ ca-c+b\leqslant ca+b^2-cb-a^2\Rightarrow cb+a^2+b\leqslant c+b^2. $$On the other hand, $cb>b^2$ and $a^2+b\geqslant a+b>c$, implying that $cb+a^2+b>c+b^2$. This yields a contradiction.

We claim the following inequality: $$\frac{b-1}{a} < \frac{c(b-1)+a}{c(a-1)+b} < \frac{b}{a}$$which will imply that even $a\cdot\tfrac{c(b-1)+a}{c(a-1)+b}$ cannot be an integer. To prove the inequality, we simply cross-multiply and use the problem's condition. \begin{align*} &\quad b[c(a-1)+b] - a[c(b-1)+a] \\ &= [abc - bc + b^2] - [abc - ac + a^2] \\ &= ac - bc + b^2-a^2 \\ &= (b-a)(a+b-c) \\ &> 0 \end{align*}and \begin{align*} &\quad a[c(b-1) + a] - (b-1)[c(a-1) + b] \\ &= c(b-1)(a-(a-1)) + a^2 + b^2 - b \\ &= (c-b)(b-1) + a^2 \\ &> 0 \end{align*}as claimed.

$$ca-c+b\mid cab-ca+a^2$$$$(a,b,c)=(a,a+x,a+x+y)$$$$ca-c+b|(b-a)(b+a-c)$$$$(a+x+y)(a-1)+a+x|x(a-y)$$$$ax<(a+x+y)(a-1)+a+x \le x(a-y)<ax$$This yields a contradiction.