Proof.
The implications $(1) \Rightarrow (2) \Rightarrow (3)$ are immediate. We will prove that $(3)$ implies $(1)$. It follows from assumption that $(3)$ that there is a unique morphism $f: A \rightarrow \mathbf{1}$ for which the lower right triangle in the diagram
\[ \xymatrix@R =50pt@C=50pt{ A \otimes \mathbf{1} \ar [r]^-{\operatorname{id}_ A \otimes \epsilon } \ar [d]^{\rho _{A}} & A \otimes A \ar [d]^{\operatorname{id}_ A \otimes f} \ar [dl]_{m} \\ A \ar [r]^{ \rho _{A}^{-1} } & A \otimes \mathbf{1} } \]
commutes. The upper left triangle also commutes, since $\epsilon $ is a right unit with respect to the multiplication $m$. It follows that the square commutes: that is, the composition
\[ A \otimes \mathbf{1} \xrightarrow { \operatorname{id}_{A} \otimes \epsilon } A \otimes A \xrightarrow { \operatorname{id}_{A} \otimes f } A \otimes \mathbf{1} \]
is equal to the identity. Invoking assumption $(3)$, we conclude that $f$ is a left inverse to $\epsilon $: that is, the composition $f \circ \epsilon $ is equal to the identity on the unit object $\mathbf{1}$.
We now show that $f$ is also a right inverse to $\epsilon $: that is, the composition $\epsilon \circ f$ is equal to the identity morphism $\operatorname{id}_{A}$. Consider the diagram
\[ \xymatrix@R =50pt@C=50pt{ A \ar [r]^{ \lambda _{A}^{-1} } \ar [dd]^{f} & \mathbf{1} \otimes A \ar [r]^{ \epsilon \otimes \operatorname{id}_{A} } \ar [d]^{ \operatorname{id}_{\mathbf{1}} \otimes f } & A \otimes A \ar [d]^{ \operatorname{id}_{A} \otimes f} \\ & \mathbf{1} \otimes \mathbf{1} \ar [r]^{ \epsilon \otimes \operatorname{id}_{\mathbf{1} } } \ar [dl]_{\upsilon } & A \otimes \mathbf{1} \ar [d]^{ \rho _{A} } \\ \mathbf{1} \ar [rr]^{\epsilon } & & A. } \]
The defining property of $f$ guarantees that the vertical composition on the right coincides with the multiplication map $m: A \otimes A \rightarrow A$. The assumption that $\epsilon $ is a left unit with respect to the multiplication $m$ shows that clockwise composition around the diagram gives the identity map $\operatorname{id}_{A}: A \rightarrow A$. To complete the proof, it will suffice to show that the diagram commutes. The commutativity of the upper right square follows from the functoriality of the tensor product, the commutativity of the trapezoidal region on the left follows from the functoriality of the left unit constraints of $\operatorname{\mathcal{C}}$, and the commutativity of the trapezoidal region on the bottom from the functoriality of the right unit constraints of $\operatorname{\mathcal{C}}$ (here we invoke the fact that the map $\upsilon : \mathbf{1} \otimes \mathbf{1} \xrightarrow {\sim } \mathbf{1}$ coincides with both $\lambda _{\mathbf{1}}$ and $\rho _{\mathbf{1}}$; see Corollary 2.1.2.21).
$\square$