Let $p_X(X,\theta)=\theta(1-\theta)^{k-1}$ where $k\in\mathbb{N}$ and $0<\theta<1$.
$L(\theta)=\prod_{i=1}^{n} p_X=\prod_{i=1}^{n} \theta(1-\theta)^{k-1}=\theta^n(1-\theta)^{\sum k_i-1}=\theta^n(1-\theta)^{n-\sum k_i}$
Taking the natural log of both sides:
$n\ln(\theta)+(n-\sum k_i)\ln(1-\theta)$
Taking the derivative w.r.t $\theta$ and setting it to $0$:
$0=\frac{n}{\theta}-\left(\frac{n-\sum k_i}{1-\theta}\right)$
Solving this is where I'm getting some issues
$\frac{n-\sum k_i}{1-\theta}=\frac{n}{\theta}$
$1-\bar{X}=\frac{1-\theta}{\theta}$
I'm not sure where to go from here.
The problem is an algebra error in your likelihood function. $\Sigma(k_i-1)=(\Sigma k_i)-n$, not $n-\Sigma k_i$.
Something to consider is that when you solve the last expression for theta you get $1/(2-\bar{x})$, which isn't restricted to (0,1) since the pmf is defined for $X=1,2,3,...$; in fact, it isn't even defined if $\bar{x}=2$! If you get an impossible answer you should go back and check the rest of your work.