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 http://mathoverflow.net/questions/249252/ext1-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 $\text{Ext}^1(M,N_1)=0$.

More generally, if $\alpha: M^m\to M^n$ ($m>0$) is a map between direct sums of copies of $M$ all of whose components are in the radical of $\text{End}(M)$, and $F$ is an additive functor such that $F(\alpha)$ is injective, then $F(M)=0$. Without loss of generality, $n$ is a multiple of $m$, as otherwise we could add extra summands to $M^n$ until it is. Say $n=dm$. Splitting $M^{dm}$ into $d$ summands, and mapping each to $M^{dm}$ by $\alpha$, we get a map $M^{dm}\to M^{d^2m}$. Continuing in this way, we get a sequence of maps
$$M^m\to M^{dm}\to M^{d^2m}\to\dots\to M^{d^km},$$
all of which become injective upon applying $F$. But for $k$ greater than the Loewy length of $\text{End}(M)$ the composition of this sequence of maps is zero, so $F(M^m)=0$.