Finding eigenvalues of an unknown matrix subtracted by the identity matrix

$x$ is an eigenvalue if $det(A - xI) = 0$. Rewrite slightly to get $$ det((A-I) - (x-1)I) = 0 $$ or $$ det((A-I) - yI) = 0. $$

So: a solution $y$ to the last equation is an eigenvalue of $A-I$. But in that case, $x = y+1$ is an eigenvalue of $A$. So your conjecture is correct.

For any matrix $B$, we have that $\lambda$ is an eigenvalue of $B$ with eigenvector $v$ if and only if, for all scalars $\mu$, $\lambda + \mu$ is an eigenvalue of $B + \mu I$ with the same eigenvector, since

$Bv = \lambda v \tag{1}$

is clearly equivalent to

$(B + \mu I)v = Bv + \mu I v = (\lambda + \mu)v. \tag{2}$

Applying this equivalence to the case at hand shows that the eigenvalues of $A$ are indeed $-1$, $0$, and $2$.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!