Let $A$ be a local selfinjective algebra with indecomposable module $M$.
Let $N=A \oplus M$. 


When there is an indecomposable module $U$ not in $add(N)$, having finite $add(N)$-resolution for some choice of $A$ and $M$? 


This is not possible in general due to the following examples:


* $A=K[x]/(x^n)$ for arbitary $M$
* $A$ arbitrary and $M$ simple.



Finite add(N)-resolution
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](https://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 $F(M)=0$. 


To see this, let $E=\text{End}(M)$, and suppose $\beta: M\to M$ is in the radical of $E$. If $F(M)\neq0$ then choose $x\neq0$ in $\text{rad}^s(E)F(M)$ where $s$ is maximal subject to $\text{rad}^s(E)F(M)\neq0$. Then $F(\beta)(x)=\beta x\in\text{rad}^{s+1}(E)F(M)=0$. Therefore the element of $F(M^n)=F(M)^n$ with all components equal to $x$ is a nonzero element of $\ker(\alpha)$.


[This is the sillier proof of the last claim that I originally posted:


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 $F(M^m)=0$.]


