If $2p+1$ is composite there is a solution with $k=1$ and if it is prime then there is none with $k=1$. In general there is a solution with $k=K$ if and only if $(K+1)p+K^2$ factors as the product of two factors both congruent to -1 $\mod K$. I suspect that there are infinitely many exceptions however this would imply that that there are infinitely many Sophie Germain primes (2p+1 also prime). This is not known to be true although there is every reason to expect that there are about $\frac{n}{\log^2n}$ such less than $n$.

$p=(k+1)(ab)+k(a+b)$ is true exactly if $(k+1)p+k^2=((k+1)a+k)((k+1)b+k)$. So the number $(k+1)p+k^2=(k+1)(p+k-1)+1$ must factor as the product of two numbers congruent to $-1 \mod k+1$

The largest that $k$ could be is $\frac{p-2}{5}$ in case $(a,b)=(1,1)$. Of course then $2p+1$ is not prime (if $p \ne 2$)

It may be that there are only finitely many primes with $(k+1)p+k^2$ prime for all $k$ up to $p/5$ however we don't need them all prime, just to not have any factor congruent to $k \mod k+1.$

The first few Sophie Germain primes are
$$2, 3, 5, 11, 23, 29, 41, 53, \mathbf{ 83}, 89, 113, 131, 173, 179, 191, 233, 239, 251, \mathbf {281}, 293,\mathbf{359}$$ None of these satisfy an equivalence as above *except* 83 for $(a,b)=(2,3),(2,7),(3,7)$ ,281 for $(a,b)=(2,3),(4,5),(4,9),(5,9),(2,25),(3,25)$ and 359 for $(a,b)=(2,7),(2,15),(7,15)$

In all there are 65 primes under 10000 which do not satisfy an equivalence of the desired type. All of the exceptions are Sophie Germain primes. The largest Sophie Germain primes under 1000 are

$$9371, \mathbf{9419, 9473, 9479}, 9539, \mathbf{9629, 9689, 9791}$$ and all of these DO satisfy a congruence of the desired type with the exceptions of 9371 and 9539.