Thanks for the hint, Erick.
I now came up with the following argument. Suppose $A$ is a real positive definite symmetric Matrix and that $B$ is a real symmetric logarithm of $A$, i.e. $B$ is real and symmetric and $\exp(B) = A$.
We wish to show that $B$ is uniquely determined by $A$. Since $B$ is diagonalizable, it suffices to show that $\sigma(B)$, the set of eigenvalues of $B$, is uniquely determined by $A$, and that every eigenspace $E_t(B) = \lbrace v \in \mathbb{R}^n \, : \, Bv = tv \rbrace$, $t \in \sigma(B)$ is uniquely determined by $A$.
Since $B$ is symmetric, $B$ is similar to a diagonal matrix $D$, which has entries from $\sigma(B)$. Hence $\exp(B) $ is similar to $\exp(D)$. It follows that
$$\sigma(\exp(B)) = \lbrace e^t \, : \, t \in \sigma(B) \rbrace \Leftrightarrow \sigma(B ) = \lbrace \log(s) \,: \, s \in \sigma(A) \rbrace$$
Since $t \mapsto e^t$ is a bijection from $\mathbb{R}$ to $(0, \infty)$.
We next show that for every eigenvalue $t$ of $B$ we have
$$E_{t}(B) = E_{e^t}(\exp(B))$$
Let $v \in E_{t}(B)$. Then
$$\exp(B)v = \sum_{k\geq 0}(\frac{B^k}{k!})v = \sum_{k\geq 0}(\frac{B^kv}{k!}) = \sum_{k\geq 0}(\frac{t^kv}{k!}) = e^t v$$
Hence $E_t(B) \subset E_{e^t}(\exp(B))$. In order to prove equality, we prove that the dimensions of both spaces are equal. Since both $B$ and $\exp(B)$ are diagonalizable, we have the direct sum decompositions
$$ \bigoplus_{t \in \sigma(B)} {E_t(B)} = \mathbb{R}^n = \bigoplus_{t \in \sigma(B)} {E_{e^t}(\exp(B))}$$
Hence
$$\sum_{t \in \sigma(B)} {\dim(E_t(B))} = n = \sum_{t \in \sigma(B)} {\dim(E_{e^t}(\exp(B)))}$$
This implies $\dim(E_t(B)) = \dim(E_{e^t}(\exp(B)))$ for every $t \in \sigma(B)$, as desired.
