Does this computation lead to solving DL? Imagine we have $g^x$, $a$ and $g^b \in \mathbb{Z}_p$. Is it realistic to compute $g^{(a+x)b}$ without knowing $x$ and $b$? Or is it equivalent to solving the Discrete Logarithm problem?
No.
Assuming that uniformly random sampling for $x,b$ is applied, this (let's call it E-CDH) is actually equivalent (in terms of hardness) to the Computational Diffie-Hellman Problem (CDH). Which states: Given a finite group $G$, sample $a,b\in G$ uniformly at random and provide $g^a,g^b$. The problem is solved if you can find $g^{ab}$.

Proof:

CDH is at most as hard as E-CDH ($\Rightarrow$). Suppose we have an oracle $\mathcal O(g^b,a,g^x)$ that returns $g^{(a+x)b}$ (i.e. an oracle that solves E-CDH). We can now solve any given CDH instance $g^a,g^b$ as the result of $\mathcal O(g^a,0,g^b)=g^{(a+0)b}=g^{ab}$.

E-CDH is at most as hard as CDH ($\Leftarrow$). Suppose we have an oracle $\mathcal O(g^a,g^b)$that returns $g^{ab}$, ie solves the CDH problem. Now suppose we are given the instance $g^x,a,g^b$, we now compute $g^a$ and $g^x\cdot g^a=g^{x+a}$. Finally we return $\mathcal O(g^{x+a},g^b)=g^{(x+a)b}$ and thus solve the instance.



Small assumption: We need to actually know $g$ and the relevant group.