Clearly if $f(n) = n - 1$ for $n > 1$, the desired identity will be satisfied. In fact one can easily check that the value of $f(2)$ is also irrelevant, so any $f$ such that $f(n) = n - 1$ for all $n > 2$ will work. We show that these are the only such functions.
Plug in $a = b = c$ to obtain $f^{a^3-a}(a^3) = a$ for all $a$. Using this twice, $f^{a^9-a}(a^9) = a$. Then, $b = c = a^4$ gives $f^{a^9-a^4}(a^9) = a^4$. Meanwhile $b = a^3, c = a^5$ gives $f^{a^9-a^5}(a^9) = a^5$; consequently, we get $f^{a^5-a^4}(a^5) = a^4$.

$$f^{bc-b}(bc) + f^{bc-c}(bc) = b + c. \qquad (19)$$
Now fix a large number $N$, and for all divisors $d \mid N$ with $1 < d < N$, define $g(d)$ by $g(d) = f^{N-d}(N)-d$. We claim that $g(bc) = g(b) + g(c)$ whenever $bc$ is still a proper divisor of $N$. Proof: put $a = N/bc$, and write
$$\begin{aligned} g(bc) &= f^{abc-bc}(abc) - bc \\ &= (a + bc - f^{abc-a}(abc)) - bc \quad (\text{by (19)}) \\ &= a - f^{abc-a}(abc) \\ &= f^{abc-b}(abc) - b + f^{abc-c}(abc) - c \\ &= g(b) + g(c). \end{aligned}$$
This holds for any $N$. In particular, fix any $b \ge 2$, and put $N = (b^r(b+1)^s)^3$, where $r, s$ are any relatively prime integers both greater than $b+1$. Repeatedly using the multiplicative property of $g$ defined above, we get
$$rg(b) + sg(b+1) = g(b^r(b+1)^s) = 0.$$
The only solution to this with integers $g(b), g(b+1)$ satisfying $g(b) \ge -b$ and $g(b+1) \ge -(b+1)$ is $g(b) = g(b+1) = 0$. Hence,
$$f^{N-b}(N) = b, \quad f^{N-(b+1)}(N) = b+1$$
and comparing gives $f(b+1) = b$, which is what we set out to prove.