There is a variational criterion to determine whether geodesic balls are convex in finite dimensional open Alexandrov space $(X,d)$. Otherwise, if $X$ is compact, then my answer says nothing.

If $X$ is connected, let $M^*=M^*(x_0)$ be the set of noncompact maximal minimizing geodesic rays. For $\lambda \in M^*$, let $h_\lambda(x)$ be unique horofunction centred at $\lambda$ and satisfying $h_\lambda(x_0)=0$. Then $$b(x,y):=\sup_{\lambda \in M^*} |h_\lambda(x)-h_\lambda(y)|$$ defines a possibly degenerate metric on $b:X\times X\to [0,+\infty)$ satisfying $$b(x,y)\leq d(x,y)$$ for every $x,y \in X$. From the $d$-geodesic convexity of $h_\lambda$, we find the $b$-metric balls of arbitrary radius are $d$-geodesically convex subsets of $X$. The $d$ geodesic convexity of the horofunction $h_\lambda$ is well-known property of Alexandrov metrics $d$.

If $(X,d)$ has the property that equality is obtained everywhere $b(x,y)=d(x,y)$ we conclude all $d$-metric balls in $X$ are $d$-geodesically convex.

So in your case, we need further information on the set of infinite minimizing rays $M^*=M^*(x_0)$ at a basepoint $x_0$. In plain terms, for every pair of points $x,y$, you need determine whether there exists a horosphere $H_\lambda$ which separates $x,y$. In such case $b$ is nondegenerate distance function with $b(x,y)=0$ iff $x=y$.

If equality is attained everywhere and $b=d$, then $b$ is Alexandrov. But if the inequality is strict $b<d$ , then I don't know whether the metric is again Alexandrov, and this appears interesting question.