I stumbled across an identity and I don't know to prove it

Question

I stumbled across this apparent identity: $$aa^{-1}_{b}+bb^{-1}_{a}-1=ab$$ where $$ a,b\in \mathbb{N/\{0\}} \\ gcd(a,b)=1 \\ (a^{-1}_{b})\equiv a^{\phi(b)-1} \mod b \\ (b^{-1}_{a})\equiv b^{\phi(a)-1} \mod a \\ $$ Pretty much $a^{-1}_{b}$ and $b^{-1}_{a}$ are multiplicative inverses and I chose for simplicity that a and b are coprime. If I run values in python it seems to be true. What I've done so far is prove that the lhs is a multiple of the rhs: $$ a^{-1}_{b} = a^{\phi(b)-1} \\ b^{-1}_{a} = b^{\phi(a)-1} \\ E = aa^{-1}_{b}+bb^{-1}_{a}-1=a^{\phi(b)}+b^{\phi(a)}-1 \\ E \equiv a^{\phi(b)}-1\equiv 0\mod b \\ E \equiv b^{\phi(a)}-1\equiv 0\mod a $$ therefore $E$ is a multiple of $a$ and a multiple of $b$ so it is a multiple of $ab$. But I can't figure out how to show that it is equal to $ab$ and not just a multiple.