As explained by Mohan in the comments, the answer is Yes.
Indeed, let $K$ be a field, let $x$ and $y$ be indeterminates, set $A:=K[x,y]$, and denote abusively by $K$ the $A$-module $A/(x,y)$. As we have
$$
\operatorname{Ext}_A^1((x,y),K)\simeq\operatorname{Ext}_A^2(K,K)\simeq K,
$$
there is a non-split exact sequence
$$
0\to K\to M\xrightarrow\pi(x,y)\to0.
$$
Clearly $K$ is the torsion module $T(M)$ of $M$, and $M$ is neither torsion nor torsion-free. It suffices to check that $M$ is indecomposable.
Let $M_1$ and $M_2$ be two submodules of $M$ such that $M=M_1\oplus M_2$. We have
$$
K=T(M)=T(M_1)\oplus T(M_2),
$$
and thus $K$ is contained in $M_1$ or $M_2$. Say that $K$ is contained in $M_1$. The exact sequence being non-split, $K$ is a proper submodule of $M_1$. We get $(x,y)=\pi(M_1)\oplus\pi(M_2)$ and $\pi(M_1)\ne0$. The fact that $(x,y)$ is indecomposable implies $\pi(M_1)=(x,y)$ and thus $M_1=M$, showing that $M$ is indecomposable.