I don't think this is true, even for stable bundles. Assume that the genus of $X$ is at least 2. Take a point $p$ of $X$, and a nontrivial extension of ${\cal O}(p)$ by ${\cal O}$, this is indecomposable and stable. Using the fact that the homomorphism ${\rm H}^1(X, {\cal O}(-p)) \to {\rm H}^1(X, {\cal O})$ induced by the embedding ${\cal O}(-p) \subseteq {\cal O}$ is an isomorphism, it is easily seen that ${\rm H}^0(X, E) = k$; so there is a unique embedding $\mathcal {O} \subseteq E$. This implies that if $E'$ is another such extension, then $E$ and $E'$ are isomorphic if and only if they differ by multiplication by a nonzero scalar in ${\rm Ext}^1({\cal O(p), O})$. There is a line in ${\rm H}^1(X, \cal O(-p))$ parametrizing non-trivial extensions of $\cal O(p)$ by $\cal O$, no two of which are isomorphic. These are all specializations of the one corresponding to the generic point of the line. Since they are all indecomposable, this gives a counterexample.

I don't think this is true, even for stable bundles. Assume that the genus of $X$ is at least 2. Take a point $p$ of $X$, and a nontrivial extension of ${\cal O}(p)$ by ${\cal O}$, this is indecomposable and stable. Using the fact that the homomorphism ${\rm Ext}^1({\cal O(p), O}) = {\rm H}^1(X, {\cal O}(-p)) \to {\rm H}^1(X, {\cal O}) = {\rm Ext}^1({\cal O, O})$ induced by the embedding ${\cal O}(-p) \subseteq {\cal O}$ is an isomorphism, it is easily seen that ${\rm H}^0(X, E) = k$; so there is a unique embedding $\mathcal {O} \subseteq E$. This implies that if $E'$ is another such extension, then $E$ and $E'$ are isomorphic if and only if they differ by multiplication by a nonzero scalar in ${\rm Ext}^1({\cal O(p), O})$. There is a line in ${\rm H}^1(X, \cal O(-p))$ parametrizing non-trivial extensions of $\cal O(p)$ by $\cal O$, no two of which are isomorphic. These are all specializations of the one corresponding to the generic point of the line. Since they are all indecomposable, this gives a counterexample.

I don't think this is true, even for stable bundles. Assume that the genus of $X$ is at least 2. Take a point $p$ of $X$; the morphism ${\rm H}^1(X, {\cal O}(-p)) \to {\rm H}^1(X, {\cal O})$ induced by the embedding ${\cal O}(-p) \subseteq {\cal O}$ has a 1-dimensional kernel. Take an extension of ${\cal O}(p)$ by ${\cal O}$, whose class in ${\rm Ext}^1({\cal O(p), O}) = {\rm H}^1(X, {\cal O}(-p))$ has a nonzero image in ${\rm Ext}^1({\cal O, O}) = {\rm H}^1(X, {\cal O})$. This is indecomposable and stable. It is seen that ${\rm H}^0(X, E) = k$; so there is a unique embedding $\mathcal {O} \subseteq E$. This implies that if $E'$ is another such extension, then $E$ and $E'$ are isomorphic if and only if they differ by multiplication by a nonzero scalar in ${\rm Ext}^1({\cal O(p), O})$. There is a line in ${\rm H}^1(X, \cal O(-p))$ parametrizing non-trivial extensions of $\cal O(p)$ by $\cal O$, no two of which are isomorphic. These are all specializations of the one corresponding to the generic point of the line. Since they are all indecomposable, this gives a counterexample.

I don't think this is true, even for stable bundles. Assume that the genus of $X$ is at least 2. Take a point $p$ of $X$, and a nontrivial extension of ${\cal O}(p)$ by ${\cal O}$, this is indecomposable and stable. Using the fact that the homomorphism ${\rm H}^1(X, {\cal O}(-p)) \to {\rm H}^1(X, {\cal O})$ induced by the embedding ${\cal O}(-p) \subseteq {\cal O}$ is an isomorphism, it is easily seen that ${\rm H}^0(X, E) = k$; so there is a unique embedding $\mathcal {O} \subseteq E$. This implies that if $E'$ is another such extension, then $E$ and $E'$ are isomorphic if and only if they differ by multiplication by a nonzero scalar in ${\rm Ext}^1({\cal O(p), O})$. There is a line in ${\rm H}^1(X, \cal O(-p))$ parametrizing non-trivial extensions of $\cal O(p)$ by $\cal O$, no two of which are isomorphic. These are all specializations of the one corresponding to the generic point of the line. Since they are all indecomposable, this gives a counterexample.