Plugging $(a, b) \leftarrow (1, 1)$ into the first given condition, we get $f(1, 1) = 1$.

On the other hand, by plugging $(a, b) \leftarrow (a, b+1)$ we get
$$f(a, b + 1) + a + b + 1 = f(a, 1) + f(1, b + 1) + ab + a.$$
Using the first given condition in its original form, it follows that
$$f(a, b + 1) - f(a, b) = f(1, b + 1) - f(1, b) + a - 1.$$
Let $p > 2$ be a fixed prime number such that $p \mid a+b$, meaning that $p \mid f(a,b)$. Since $p \mid a+(b+1)-1$, we also have $p \mid f(a,b+1)$. Therefore, $p \mid f(1,b+1)-f(1,b)+a-1$ and
$$p \mid f(1, b + 1) - f(1, b) - b - 1 \quad (*)$$
for all positive integers $a$ and $b$, and primes $p > 2$ such that $p \mid a+b$.
Note that the right-hand side in $(*)$ does not depend on $a$, hence fixing $b$ and varying $a$ would yield infinitely many odd prime divisors of $f(1, b+1) - f(1, b) - b - 1$, which is possible only if the latter equals 0, i.e.
$$f(1, b + 1) = f(1, b) + b + 1$$
for every positive integer $b$.
Therefore, $f(1, 2) = f(1, 1) + 2 = 1 + 2$, $f(1, 3) = f(1, 2) + 3 = 1 + 2 + 3$, i.e. $f(1, n) = 1 + \cdots + n = \frac{n(n+1)}{2}$ (by mathematical induction).
In a similar fashion, starting with $(a, b) \leftarrow (a+1, b)$, we infer that $f(n, 1) = \frac{n(n+1)}{2}$.
Finally, from
$$\begin{aligned} f(a, b) + a + b &= f(a, 1) + f(1, b) + ab \\ &= \frac{a(a + 1)}{2} + \frac{b(b + 1)}{2} + ab, \end{aligned}$$
it follows that $f(a, b) = \binom{a+b}{2}$. This function clearly satisfies the second given condition, hence it is the only solution.