Plugging in $k = n = m = 1$ yields
$$f(1)^2 - 2f(1) + 1 = (f(1) - 1)^2 \leq 0,$$
which implies $f(1) = 1$. Plugging in $k = 1, n = m$ and $k = n, m = 1$, respectively, we obtain the two inequalities
$$2f(n) - f(n^2) \geq 1, \qquad (1)$$
$$f(n^2) + f(n) - f(n)^2 \geq 1. \qquad (2)$$

We add the two to get
$$-f(n)^2 + 3f(n) \geq 2$$
or
$$(f(n) - 1)(f(n) - 2) \leq 0.$$
This means that $1 \leq f(n) \leq 2$ for all $n$. Now assume that there is a positive integer $n$ such that $f(n) = a > 1$. The function $g(x) = \frac{x^2+1}{x} = x + \frac{1}{x}$ is increasing on $(1, \infty)$: for $x > y > 1$,
we have
$$g(x) - g(y) = x - y + \frac{1}{x} - \frac{1}{y} = (x - y) \left(1 - \frac{1}{xy}\right) > 0.$$
Hence, for any $x \geq a$, we have
$$x^2 + 1 \geq \frac{a^2 + 1}{a}x$$
and thus
$$x^2 - x + 1 \geq \frac{a^2 - a + 1}{a} \cdot x.$$
Returning to (2), we now find, by our assumption that $f(n) = a$,
$$f(n^2) \geq f(n)^2 - f(n) + 1 \geq \frac{a^2 - a + 1}{a} \cdot f(n),$$
and since
$$b = \frac{a^2 - a + 1}{a} = 1 + \frac{(a - 1)^2}{a} > 1,$$
we get
$$f(n^2) \geq b f(n) \geq a.$$
Iterating this inequality yields $f(n^4) \geq b^2 f(n) = b^2 a$, $f(n^8) \geq b^3 f(n) = b^3 a$, etc., and generally (by induction) $f(n^{2^k}) \geq b^k a$. Since $a > 1$ and $b > 1$, this implies $f(n^{2^k}) > 2$ for sufficiently large $k$, which contradicts the inequality $1 \leq f(n^{2^k}) \leq 2$ that was obtained earlier.
It follows that there is no $n$ such that $f(n) > 1$, which means that the constant function $f(n) \equiv 1$ is the only solution (and it is easy to see that this function satisfies the condition, since the left hand side of the inequality is always equal to 1 in this case).

It is given that
$$f(km) + f(kn) - f(k)f(mn) \geq 1 \quad (3)$$
for all $k, m, n \in \mathbb{N}$.
Put $m = n = k = 1$ into (3) to get $(f(1) - 1)^2 \leq 0$, giving
$$f(1) = 1. \quad (4)$$
Put $m = n = 1$ into (3) to get $2f(k) - f(k) \geq 1$, giving
$$f(k) \geq 1 \quad (5)$$
for all $k \in \mathbb{N}$. Put $k = m = n$ into (3) to get $2f(k^2) - f(k)f(k^2) \ge 1$, giving
$$f(k^2)(2 - f(k)) \ge 1, \qquad (6)$$
again for all $k \in \mathbb{N}$. From (5) and (6) it follows that
$$1 \le f(k) < 2 \qquad (7)$$
for all $k \in \mathbb{N}$. We now prove the statement
$$S(M) : 1 \le f(k) < 1 + \frac{1}{M} \text{ for all } k \in \mathbb{N}$$
to be true for all $M \ge 1$, using induction on $M$. The case $M = 1$ is just (7).
Assume $S(M)$ to be true. Suppose there is a $k_0 \in \mathbb{N}$ such that $f(k_0) \ge 1 + \frac{1}{M+1}$. Then, using (6), we obtain
$$1 \le f(k_0^2)(2 - f(k_0)) \le f(k_0^2) \left(2 - 1 - \frac{1}{M+1}\right) = f(k_0^2) \left(\frac{M}{M+1}\right),$$
giving $f(k_0^2) \ge 1 + \frac{1}{M}$, which violates $S(M)$. So $S(M+1)$ follows.
It is now clear that the constant function $f(n) \equiv 1$ is the only solution.