L. Schwartz and A. Grothendieck made clear, by very early 1950s, that the Cauchy (-Goursat) theory of holomorphic functions of a single complex variable extended with essentially no change to functions with values in a locally convex, quasi-complete topological vector space. Cauchy integral formulas, residues, Laurent expansions, etc., all succeed (with trivial modifications occasionally).
Conceivably one needs a little care about the notion of "integral". The Gelfand-Pettis "weak" integral suffices, but/and a Bochner version of "strong" integral is also available.
Further, in great generality, as Grothendieck made clear, "weak holomorphy" (that is, $\lambda\circ f$ holomorphic for all (continuous) linear functionals $\lambda$ on the TVS) implies ("strong") holomorphy (i.e., of the TVS-valued $f$).
(Several aspects of this, and supporting matter, are on-line at http://www.math.umn.edu/~garrett/m/fun/Notes/09_vv_holo.pdf and other notes nearby on http://www.math.umn.edu/~garrett/m/fun/)
Edit: in response to @Christopher A. Wong's further question... I've not made much of a survey of recent texts to see whether holomorphic TVS-valued functions are much discussed, but I would suspect that the main mention occurs in the setting of resolvents of operators on Hilbert and Banach spaces, abstracted just a little in abstract discussions of $C^*$ algebras. (Rudin's "Functional Analysis" mentions weak integrals and weak/strong holomorphy and then doesn't use them much, for example.) Schwartz' original book did treat such things, and was the implied context for the first volume of the Gelfand-Graev-etal "Generalized Functions". In the latter, the examples are very small and tangible, but (to my taste) tremendously illuminating about families of distributions.
Edit-edit: @barto's further question is about the behavior of holomorphic operator-valued $f(z)$ at an isolated point $z_o$ where $f(z)$ fails to be invertible. I do not claim to have a definitive answer to this, but only to suggest that the answer may be complicated, since already for the case $f(z)=(T-z)^{-1}$ for bounded, self-adjoint $T$, it seems to take a bit of work (the spectral theorem) to prove that isolated singularities are in the discrete/point spectrum of $T$. But this may be overkill, anyway...
Morera's theorem, when properly(1) stated, is indeed the exact converse of Goursat's theorem.
Theorem (Morera): Let $\Omega\subset\mathbb{C}$ open, and $f\colon\Omega\to\mathbb{C}$ a continuous function. If for all triangles $T\subset\Omega$ whose interior is also contained in $\Omega$ $$\int_T f(z)\,dz = 0,$$ then $f$ is holomorphic in $\Omega$.
Instead of triangles, one could of course also use rectangles, or other polygons. And actually, we could drop the condition that the interior of the triangle be contained in $\Omega$ and be left with a still true, but arguably less useful result, since we would then have a sufficient but not necessary condition (consider $1/z$ on $\mathbb{C}\setminus \{0\}$ to have function satisfying the condition as stated, but not the stronger condition one obtains by dropping "whose interior is also contained in $\Omega$").
The condition entails the existence of local primitives of $f$, i.e. every $z\in \Omega$ has a neighbourhood $U$ such that $f = F'$ for a holomorphic function $F$ on $U$. Thus $f$ is holomorphic on $U$ (the derivative of a holomorphic function is again holomorphic), and since holomorphicity is a local property, $f$ is holomorphic on $\Omega$.
To establish the existence of local primitives, one considers (for example) for $z_0 \in \Omega$ a disk $U = D_r(z_0) \subset \Omega$, and on $U$ the function $F(z) = \int_{z_0}^z f(\zeta)\,d\zeta$. The vanishing of the integral of $f$ over triangles whose interior is contained in $\Omega$ then yields $F(z) - F(w) = \int_w^z f(\zeta)\,d\zeta$, from which $F' = f$ follows with the continuity of $f$.
(1) The term "properly" means "properly for this purpose", or "adequately to show it is the converse of Goursat's theorem" here. Stating it for simply connected domains or disks is not wrong.
There is one downside to stating it explicitly for simply connected domains, however. Often, people aren't aware of the local character of the theorem, and consider the simple connectedness as essential for the validity of the theorem. The essential point is the locality, that one considers not the entire domain $\Omega$, but a small convex neighbourhood $U\subset \Omega$ of a point $z\in \Omega$ to construct the local primitive.
Best Answer
Once you understand the basic trick of using Hahn-Banach to reduce everything to the situation where the target space is $\Bbb{C}$, I'm sure you can easily extend several results by yourself. What I did a few months back (and what I recommend you do) is pick your favorite complex analysis text (e.g I took Cartan), and I went through each of the basic theorem one-by one and extended them to the case where the target is an arbitrary Banach space, and the domain is an open subset of $\Bbb{C}^n$. Usually such extensions aren't too hard. Of course, from time to time I would refer to Mujica's text to see if I was missing out any details in the argument.
As an example of how Hahn-Banach can be used here, I shall prove Cauchy's integral formula:
Just so we're clear, by holomorphic on $U$, I mean once-complex differentiable at every point of $U$ ($X$ is a Banach space, so the difference quotient makes sense and limits make sense). Also, on the RHS, I'm referring to the Bochner integral of a continuous function $[a,b]\to X$, namely $\frac{1}{2\pi i}\int_a^b\frac{f(\gamma(t))}{\gamma(t)-z}\,\gamma'(t)\,dt$.
We prove this by letting $\lambda\in X^*$ be arbitrary (i.e a continuous linear map $X\to\Bbb{C}$). Then, note that $\lambda\circ f:U\to\Bbb{C}$ is holomorphic (chain rule), so by the one-dimensional Cauchy-integral formula, \begin{align} \lambda\bigg(I(\gamma,z)f(z)\bigg)&=I(\gamma,z)(\lambda\circ f)(z)\\ &=\frac{1}{2\pi i}\int_{\gamma}\frac{(\lambda\circ f)(\zeta)}{\zeta-z}\,d\zeta\tag{by 1D version}\\ &=\frac{1}{2\pi i}\int_{\gamma}\lambda\left(\frac{f(\zeta)}{\zeta-z}\right)\,d\zeta\tag{since $\lambda$ is linear}\\ &=\lambda\left(\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{\zeta-z}\,d\zeta\right) \end{align} In the last line, I used one of the basic properties of Bochner integrals, that for a Bochner-integrable function $\phi:S\to X$ from a measure space $S$ to a Banach space $X$, and $\lambda\in X^*$, we have $\lambda(\int_S\phi\,d\mu)=\int_S(\lambda\circ \phi)\,d\mu$ (in our case, you may wish to write out $\int_{\gamma}$ as an integral over $[a,b]$ to see that I'm using the theorem with $S=[a,b]$ and $\mu$ being Lebesgue measure).
Now, since $\lambda\in X^*$ was arbitrary, the Hahn-Banach theorem tells us the two things inside of $\lambda$ must be equal: \begin{align} I(\gamma,z)f(z)&=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{\zeta-z}\,d\zeta \end{align}
From here of course, the fact that "holomorphic on a disc $D_R(a)$ implies analytic on the disc $D_R(a)$" follows immediately by Taylor expanding the integrand as in the single variable case (I'm being slightly sloppy here, see this answer for the correct details). We fix $0<r<R$, then (by the just-established Banach-valued case of Cauchy's integral formula) \begin{align} f(z)&=\frac{1}{2\pi i}\int_{|\zeta-a|=r}\frac{f(\zeta)}{\zeta-z}\,d\zeta\\ &=\frac{1}{2\pi i}\int_{|\zeta-a|=r}\frac{f(\zeta)}{\zeta-a}\cdot\frac{1}{1-\frac{z-a}{\zeta-a}}\,d\zeta\\ &=\frac{1}{2\pi i}\int_{|\zeta-a|=r}\frac{f(\zeta)}{\zeta-a}\sum_{n=0}^{\infty}\left(\frac{z-a}{\zeta-a}\right)^n\,d\zeta\\ &=\sum_{n=0}^{\infty}\left(\frac{1}{2\pi i}\int_{|\zeta-a|=r}\frac{f(\zeta)}{(\zeta-a)^{n+1}}\,d\zeta\right)\cdot (z-a)^n, \end{align} where, as explained in the link, the exchange of series integral is possible due to uniform convergence.
This proves "holomorphic implies analytic". The fact that "analytic implies holomorphic" is much easier; simply take your favorite complex-analysis book and look at the proof there. The fact that $f$ is Banach-valued makes no difference (just replace absolute values by norms in appropriate places).
Now that you have the equivalence of holomorphic and analytic, I doubt you'll need Goursat's lemma (but you can still prove it if you want; just use Hahn-Banach to reduce to the one-dimensional case, and then invoke the already-known version of Goursat's lemma). Likewise, you can prove Morera's theorem, and establish Cauchy's inequalities, Louiville's theorem. With Morera's theorem in the Banach-case, you can once again prove that uniform limits of holomorphic functions are holomorphic.