We can compute that \[\mathbf{M}^3 = \begin{pmatrix} a^3 + 2abc + bcd & a^2 b + abd + bd^2 + b^2 c \\ a^2 c + acd + cd^2 + bc^2 & abc + 2bcd + d^3 \end{pmatrix}.\]Hence, $ a^2 b + abd + bd^2 + b^2 c = b(a^2 + ad + d^2 + bc) = 0, $ and $ a^2 c + acd + cd^2 + bc^2 = c(a^2 + ad + d^2 + bc) = 0 $. Furthermore, \[(\det \mathbf{M})^3 = \det (\mathbf{M}^3) = \det \mathbf{I} = 1,\]so $ \det \mathbf{M} = 1 $. In other words, $ ad - bc = 1 $. From the equation $ b(a^2 + ad + bd^2 + bc) = 0, $ either $ b = 0 $ or $ a^2 + ad + d^2 + bc = 0 $. If $ b = 0, $ then \[\mathbf{M}^3 = \begin{pmatrix} a^3 & 0 \\ a^2 c + acd + cd^2 & d^3 \end{pmatrix}.\]Hence, $ a^3 = d^3 = 1, $ so $ a = d = 1, $ and $ a + d = 2 $. Also, $ c + c + c = 0, $ so $ c = 0 $. Thus, $ \mathbf{M} = \mathbf{I} $. Otherwise, $ a^2 + ad + d^2 + bc = 0 $. Since $ ad - bc = 1, $ this becomes \[a^2 + ad + d^2 + ad - 1 = 0,\]which means $ (a + d)^2 = 1 $. Either $ a + d = 1 $ or $ a + d = -1 $. Note that \begin{align*} \mathbf{M}^2 - (a + d) \mathbf{M} + (ad - bc) \mathbf{I} &= \begin{pmatrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end{pmatrix} - (a + d) \begin{pmatrix} a & b \\ c & d \end{pmatrix} + (ad - bc) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ &= \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = \mathbf{0}.\end{align*}If $ a + d = 1, $ then \[\mathbf{M}^2 - \mathbf{M} + \mathbf{I} = \mathbf{0}.\]Then $ (\mathbf{M} + \mathbf{I})(\mathbf{M}^2 - \mathbf{M} + \mathbf{I}) = \mathbf{0} $. Expanding, we get \[\mathbf{M}^3 - \mathbf{M}^2 + \mathbf{M} + \mathbf{M}^2 - \mathbf{M} + \mathbf{I} = \mathbf{0},\]which simplifies to $ \mathbf{M}^3 = -\mathbf{I} $. This is a contradiction, because $ \mathbf{M}^3 = \mathbf{I} $. Then the only possibility left is that $ a + d = -1 $. Note that \[\mathbf{M} = \begin{pmatrix} 0 & -1 \\ 1 & -1 \end{pmatrix}\]satisfies $ \mathbf{M}^3 = \mathbf{I}, $ so $ -1 $ is a possible value of $ a + d $. Thus, the only possible values of $ a + d $ are $ \boxed{2, -1} $.