Here are a few ideas on places to look for examples. You may not find (m)any this way.

First to eliminate a trivial case: For some people a complete graph or disjoint union of isomorphic complete graphs is a SRG. These only have two eigenvalues even without making any changes.

A SRG need not have any automorphisms but many do. This allows an easier complete search of very small cases. In a very modest search I found one legit example and a few questionable ones. Then it seems reasonable to look for ways to sign the matrix and preserve a large subgroup of the automorphism group (but I did not get anything from that.)

Essentially we are taking a graph with $e$ edges, viewing it as having $2e$ directed edges, and then giving some of them a weight of $-1$.

A 4-cycle has 4 edges or $8$ directed edges (aka $+1$ entries of the Adjacency matrix $A$)

- Changing a single $1$ to $-1$ does not work.
- There are $28$ ways to choose two entries, but only $6$ up to action of the Dihedral group. Changing two entries , both from the same edge, gives eigenvalues $\pm\sqrt{2}$ twice each. (That is my sole good example so enjoy it).
There are also three ways to change two entries to $-1$ and get all eigenvalues $0$:
the two edges leaving a vertex, the two going into a vertex, and two parallel directed edges.
- there are no successful ways to change three entries.
- There are $\binom84=70$ ways to choose $4$ entries to change but only $13$ up to isomorphism (maybe less but at worst I looked at some cases twice.) Four of them give all eigenvalues $0.$

Note that with two eigenvalues $\theta_1,\theta_2$ taken $k$ and $n-k$ times we need $\frac{\theta_1}{\theta2}=-\frac{n-k}{k}$ unless it is actually a single eigenvalue of $0$.

A pentagon amounts to $10$ directed edges, There is no way to change some of the entries of $A$ to $-1$ and get only two eigenvalues. There are $\binom{10}{5}=252$ ways to change $5$ of them but only $26$ up to isomorphism.

The complete bipartite graph has $18$ directed edges. Nothing works there. There would be $\binom{18}{9}=24310$ ways to change half the edges but only $681$ up to isomorphism.

I'd hoped to find an example which generalized. I did not give up after two tries but did after three. Maybe someone else will find something by looking a bit harder. Perhaps $K_{2,2,2}$ , $K_{4,4}$, or some other small case.

I also looked, without results at a few ways to weight the $2 \cdot 15=30$ directed edges from a Peterson Graph.

- The obvious order $5$ rotation gives $3$ pairs of $5$ edge orbits so $64$ (or $32$ or $16$ depending how hard you wish to think) ways to sign some orbits $-1$. None worked ( assuming I programmed correctly).
- Fixing a point gives orbits of sizes $3,3,6,6$ and $12$ (Probably the $12$ could be split $6,6$ but I did not try that variation. ) That does not yield anything.
- I did not look at fixing a pair of vertices (setwise). This would give orbits of sizes $1,1,4,4,8,8,2?,2?.$

A similar attempt could be made for other graphs.