I believe the answer to your question is "no". I am not familiar with the term "Riemannian logarithmic map," but I imagine you mean the inverse of the Riemannian exponential map. Your condition is thus that the exponential map from any point of $M$ is a diffeomorphism (in particular this implies that $M$ is diffeomorphic to $\mathbb{R}^n$).
Since the exponential map from $x$ is a diffeomorphism, there exist geodesic normal co-ordinates about $x$ for the whole manifold, and so the corresponding distance function $r=d(x,\cdot)$ is smooth on the whole manifold (except at $x$).
At each point $y$ in $M\setminus\{x\}$, one has that $\operatorname{exp}_y(-r\nabla r) =x$. So, in your terminology, $\operatorname{log}_y(x)=-r\nabla r$.
You ask essentially whether, at each point $c$ in $M$, for each vector $v$ in $T_cM$,
$$\lVert\nabla_{v}(-r\nabla r)\rVert_c=\lVert v\rVert_c;$$
equivalently whether the map from $T_cM$ to $T_cM$,
$$v\mapsto \nabla_v(-r\nabla r),$$
is always an isometry (of vector spaces). In general it is not. For example, in the hyperbolic plane, if I calculate correctly,
$$\nabla_{\frac{\partial}{\partial\theta}}(r\nabla r)=r\coth(r)\frac{\partial}{\partial\theta}.$$
This is closely related to the famous question of the "Hessian of the square of the distance function" (since $\nabla (-rdr)=-\tfrac{1}{2}\operatorname{Hess}(r^2)$), which is equal to the identity only in flat manifolds.