Generators of the $SU(2) $group Suppose we have a general unitary $2\times 2$ matrix. The condition $\hat{U} \hat{U}^\dagger = I$ imposes four constraints; therefore, we can express it in terms of four real parameters.  Equivalently, we may consider four linearly independent $2\times 2$ matrices which represent the generators of the transformation
$$\hat{U} = \exp (i \alpha_i \hat{G}_i).$$
One of the generators can be identified as
$$\hat{U} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}e^{i \phi}$$
The remaining three unitary matrices have the property $\det U =1.$ The three matrices representing the Hermitian generators of the $SU(2)$ group are linearly independent from the identity and are therefore traceless. One suitable choice of three Hermitian traceless generators are the Pauli spin-matrices.
My question is:
My textbook [Modern Particle Physics by Mark Thomson page 212] states that

"The remaining three unitary matrices form a special unitary $SU(2)$ group with the property $\det U =1.$"

But how can these form a group if they do not include the identity? I am assuming the group operation is matrix multiplication.
 A: You have the group of unitary matrices $\mathrm U(2)$, and you have the generators of that group, which constitute an algebra.  It's important not to get these things confused.
$\mathrm U(2)$ consists of all $2\times 2$ complex matrices $ U$ such that $ U^\dagger U =  U U^\dagger = \mathbb I$.  The set of such matrices forms a (non-abelian) group with the group operation being ordinary matrix multiplication.
The set of generators of this group, which we denote $\mathfrak u(2)$, consists of all $2\times 2$ complex matrices $g$ which are antihermitian, i.e. $g^\dagger = -g$.  The set of such matrices constitutes a vector space, and when we equip it with the product operation $[g,h]:= gh - hg$, it becomes an algebra (in fact, a Lie algebra).
A suitable basis for $\mathfrak u(2)$ is given by $\tau_0 = \pmatrix{i & 0 \\ 0 & i}$ together with the Pauli matrices multiplied by the imaginary unit,
$$\tau_1 = i\sigma_1 = \pmatrix{0 & i \\ i & 0} \quad \tau_2 = i\sigma_2 = \pmatrix{ 0 & 1 \\ -1 & 0} \quad \tau_3 = i\sigma_3 = \pmatrix{i & 0 \\ 0 & -i}$$
A generic element of the algebra $\mathfrak u(2)$ is some linear combination $g = \sum_n \alpha_n \tau_n = \sum_n i\alpha_n \sigma_n$, and if $g$ is exponentiated we obtain an element of $\mathrm U(2)$, namely
$$U = \exp[g] = \exp\left[\sum_n i\alpha_n \sigma_n\right]$$


One of the generators can be identified as $$
\hat U=\pmatrix{1 & 0 \\ 0 & 1}e^{i\phi}$$

This is either a typo or extremely sloppy.  This $\hat U$ is equal to $\exp[\phi \tau_0]$; the set of all matrices of this form is a subgroup of $\mathrm U(2)$ generated by the 1D algebra consisting of constant multiples of $\tau_0$.

The remaining three unitary matrices have the property $\mathrm{det}(U)=1$.

This is also not right.  The subgroup of $\mathrm{U}(2)$ which is generated by the remaining three generators $(\tau_1,\tau_2,\tau_3)$ has this property, but that's a very different statement.

But how can these form a group if they do not include the identity? I am assuming the group operation is matrix multiplication.

The group $\mathrm{SU}(2)$ does contain the identity.  It is generated by $\tau_1,\tau_2,$ and $\tau_3$, and it is these matrices which do not include $\pmatrix{1 & 0 \\ 0 & 1}$ in their span.
Altogether I strongly dislike the passage you've quoted.  The author flips back and forth between talking about the group $\mathrm U(2)$ and the algebra $\mathfrak u(2)$ which generates it, sometimes in the middle of a sentence.
