Do identical strings always have the same SHA-256 value?
Yes, if you hash the same input with the same function, you will always get the same result.

This follows from the fact that it is a hash-function. By definition a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output.

In practice there is no seed involved in evaluating a hash-function.

Now, this is how things work in practice.
On the theoretical side of things, we often talk about families of hash-functions. In that case there does exist a key that selects which member of the family we are using. The reason for this is a technical problem with the definition of collision resistance.

The naive definition of collision resistance for a single hash function $H : \{0,1\}^* \to \{0,1\}^n$ would be that for all efficient algorithms $\mathcal{A}$ the following probability is negligible $$\Pr[(x_1,x_2)\gets\mathcal{A}(1^n): H(x_1)=H(x_2)]$$

The problem with that is, that it is impossible to achieve. Given that $H$ is compressing, collisions necessarily exist. So an algorithm $\mathcal{A}$ that simply has one of those collision hardcoded and outputs it, has
$$\Pr[(x_1,x_2)\gets\mathcal{A}(1^n): H(x_1)=H(x_2)] = 1.$$
So the definition is not achievable, since this $\mathcal{A}$ by definition exists even though nobody might know what it is.

To solve this problem, we define collision resistance for a family of hash-functions $\{H_k : \{0,1\}^* \to \{0,1\}^n\}_k$.
We then define that such a family is collision resistant if it holds that the following probability is negligible
$$\Pr_{k\gets\{0,1\}^n}[(x_1,x_2)\gets\mathcal{A}(k): H_k(x_1)=H_k(x_2)].$$

Here we do not run into the same problem, because the exact function $\mathcal{A}$ needs to find a collision for is chosen uniformly at random from an exponentially large family. Since $\mathcal{A}$ could have hardcoded collisions for at most a polynomial number of functions in the family, such hash-function families are not trivially impossible.

Note that this means that there somewhat of a disconnect between the theoretical treatment of hash-functions and their practical use.