The solution you found can be shown to be unique.
The first step is to identify the constant $k$. The three rows (or, alternatively, the three columns) add up to $3k$; together, they comprise all nine digits without repetition, so $3k = 1 + 2 + \cdots + 9 = 45$, so $k = 15$.
The digit $5$ must occur in the middle. Only $5$ can participate in a three-way sum to $15$ in as many as four different ways:
$$
1+5+9 = 15 \\
2+5+8 = 15 \\
3+5+7 = 15 \\
4+5+6 = 15
$$
Since the center square participates in four such sums, it must contain the digit $5$.
$$
\begin{array}{|c|c|c|} \hline
\phantom{8} & \phantom{1} & \phantom{6} \\ \hline
\phantom{3} & 5 & \phantom{7} \\ \hline
\phantom{4} & \phantom{9} & \phantom{2} \\ \hline
\end{array}
$$
That means that the other eight digits must occur in pairs on opposite sides of the central $5$: $1$ opposite $9$, $2$ opposite $8$, $3$ opposite $7$, and $4$ opposite $6$.
Of those digits, $1$ and $9$ must occur on opposite sides, because they can each only participate in a three-way sum to $15$ in two different ways, and a corner square would be involved in three such sums. Without loss of generality, put $1$ at the center top, and $9$ at the center bottom.
$$
\begin{array}{|c|c|c|} \hline
\phantom{8} & 1 & \phantom{6} \\ \hline
\phantom{3} & 5 & \phantom{7} \\ \hline
\phantom{4} & 9 & \phantom{2} \\ \hline
\end{array}
$$
The only other sum that $1$ can be involved in is $1+6+8 = 15$. Again without loss of generality, put $8$ at upper left, and $6$ at upper right. That puts $4$ at lower left, and $2$ at lower right.
$$
\begin{array}{|c|c|c|} \hline
8 & 1 & 6 \\ \hline
\phantom{3} & 5 & \phantom{7} \\ \hline
4 & 9 & 2 \\ \hline
\end{array}
$$
That leaves room only for $3$ at center left, and $7$ at center right.
$$
\begin{array}{|c|c|c|} \hline
8 & 1 & 6 \\ \hline
3 & 5 & 7 \\ \hline
4 & 9 & 2 \\ \hline
\end{array}
$$
All other $3$-by-$3$ magic squares involving the digits $1$ through $9$ are identical to this one, up to rotation and reflection.
As to whether a solution can be found with your eight equations in nine unknowns: It requires a method more involved than solution of simultaneous equations, because that will not enforce the rule that all nine digits must be used, exactly once each.
(1) The result:
For any matrix $C=(c_{ij})\in \mathbb R^{m\times n}$, we denote
$$C_i=\sum_k c_{ik},\quad C^j=\sum_k c_{kj}.$$
Let $A=(a_{ij})\in \mathbb R^{m\times n}$, with $A_i,A^j\in \mathbb Z$. Then there exists $B=(b_{ij})\in \mathbb Z^{m\times n}$, satisfying $B_i=A_i, B^j=A^j$ and
$$|a_{ij}-b_{ij}|< 1.$$
(2)
First, by deleting those rows and columns all of whose entries are integers, we may assume each row and column contains at least two non-integer entries.
(3) Consider the bipartite graph $\Gamma$ with $r_1,\ldots, r_m,c_1,\ldots, c_n$, with $r_i\sim c_j$ i.f.f $a_{ij}\notin \mathbb Z$. We see that $\Gamma$ is a bipartite graph with minimun degree $\geq 2$ by (2). Hence $\Gamma$ contains an even cycle $$r_{i_1}c_{j_1}r_{i_2}c_{j_2}\ldots r_{i_s}c_{j_s},$$ which corresponds to a sequence of even number of non-integer entries
$$a_{i_1 j_1} a_{i_2 j_1}a_{i_2j_2}\ldots a_{i_1j_s}.$$
(4) Denote the index of entries in the cycle by $\Lambda$ and
$d(x,\mathbb Z)=\min\{x-\lfloor x\rfloor,\lfloor x\rfloor+1-x\}$.
We may assume $d(a_{i_1,j_1})=\min_{(i,j)\in \Lambda} d(a_{ij},\mathbb Z)=\epsilon$.
Then by changing the entries in the cycle consecutively with $\pm \varepsilon$, we obtain a matrix with one more integer entry than $A$.
The result follows by induction on the number of non-integer entries of $A$.
Best Answer
It becomes convenient to introduce complex numbers here.
$A^T A = AA^T = AA^* = A^*A$
so $A$ commutes with its conjugate transpose given by $A^* = A^T$, which means $A$ is normal and thus unitarily diagonalizable. So, selecting unitary matrix $U$, this implies
$U^{-1} A U =U^* A U = D$
or equivalently
$U^* A = DU^*$
your problem about row sums being equal to one implies an eigenvector equation, i.e. $A\mathbf 1 = \mathbf 1$. Your problem then wants you to confirm that $\mathbf 1^T A = \mathbf 1^T$. But this is implied by the above unitary diagonalization.
You may select (/assume WLOG) that $\mathbf u_1 = \mathbf 1$. I.e. the diagonal matrix $D$ is given by
$D = \pmatrix{1 & \mathbf 0^*\\\mathbf0& D_0}$
putting this all together
$\pmatrix{\lambda_1 \mathbf u_1^* \\*} = \pmatrix{ \mathbf 1^T \\*} = \pmatrix{1 & \mathbf 0^*\\\mathbf0& D_0}U^* = DU^* = U^* A$
thus
$\lambda_1 \mathbf u_1^* = 1 \cdot \mathbf 1^T = \mathbf 1^T = \mathbf 1^T A$
i.e. $\mathbf 1$ is a left eigenvector for $A$