I think that this is possible if and only if $\operatorname{Ext}^1_A(M,M)=0$, and so the question reduces to another question Ext^1 for a local finite dimensional selfinjective algebra that you have asked (and which I don't know the answer to).
If $\operatorname{Ext}^1(M,M)=0$ then take $U=\Omega^{-1}M$.
Conversely, if there is such a $U$, then by truncating the $\text{add}(N)$-resolution (and changing $U$) we can assume the resolution $0\to N_1\to N_0\to U\to 0$ has length $2$ and is minimal.
By minimality, $N_1$ is a direct sum of copies of $M$. By the long exact sequence of $\text{Hom}(M,-)$, the map $\text{Ext}^1(M,N_1)\to\text{Ext}^1(M,N_0)$ is injective.
So, removing the free summands from $N_0$, we have a map $\alpha:N_1\to N'_0$ between direct sums of copies of $M$, not a split injection on any summand of $N_1$, that becomes injective upon applying $\text{Ext}^1(M,-)$. But since all components of $\alpha$ are in the radical of $\text{End}(M)$, this is not possible unless $\text{Ext}^1(M,N_1)=0$.